# The Elements of Euclid for the Use of Schools and Colleges/Book VI

*BOOK VI.*

DEFINITIONS.

1. Similar rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals.

2. Reciprocal figures, namely, triangles and parallelograms, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remaining side of this other is to the remaining side of the first.

3. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment as the greater segment is to the less.

4. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base. *THEOREM*.

*Triangles and parallelograms of the same altitude are to one another as their bases.*

Let the triangles *ABC*, *ACD*, and the parallelograms *EC*, *CF* have the same altitude, namely, the perpendicular drawn from the point *A* to *BD*: as the base *BC* is to the base *CD*, so shall the triangle *ABC* be to the triangle *ACD* and the parallelogram *EC* to the parallelogram *CF'.*

Produce *BD* both ways;

take any number of straight lines *BG*, *GH*, each equal to *BC*, and any number of straight lines *DK*, *KL*, each equal to *CD*; [I. 3.

and join *AG*, *AH*, *AK*, *AL*.

Then, because *CB*, *BC*, *GH* are all equal, [*Construction*.

the triangles *ABC*, *AGB*, *AHG* are all equal. [I. 38.

Therefore whatever multiple the base *HC* is of the base *BC*, the same multiple is the triangle *AHC* of the triangle *ABC*.

For the same reason, whatever multiple the base *CL* is of the base *CD*, the same multiple is the triangle *ACL* of the triangle *ACD*.

And if the base *HC* be equal to the base *CL*, the triangle *AHC* is equal to the triangle *ACL*; and if the base *HC* be greater than the base *CL*, the triangle *AHC* is greater than the triangle *ACL*; and if less, less. [I. 38.

Therefore, since there are four magnitudes, namely, the two bases *BC*, *CD*, and the two triangles *ABC*, *ACD*; and of the base *BC*, and the triangle *ABC*, the first and the third, any equimultiples whatever have been taken, namely, the base *HC* and the triangle *AHC*; and of the base *CD* and the triangle *ACD*, the second and the fourth, any equinmltiples whatever have been taken, namely, the base *CL* and the triangle *ACL*;

and since it has been shewn that if the base *HC* be greater
than the base *CL*, the triangle *AHC* is greater than the
triangle *ACL* ; and if equal, equal ; and if less, less ;

therefore as the base *BC* is to the base *CD*, so is the
triangle *ABC* to the triangle *ACD*. [V. *Definition* 5.

And, because the parallelogram *CE* is double of the
triangle *ABC*, and the parallelogram *CF* is double of the
triangle *ACD*; [I.41.

and that magnitudes have the same ratio which their equi-
multiples have ; [V. 15.

therefore the parallelogram *EC* is to the parallelogram *CF*
as the triangle *ABC* is to the triangle *ACD*.

But it has been shewn that the triangle *ABC* is to the
triangle *ACD* as the base *BC* is to the base *CD* ;

therefore the parallelogram *EC* is to the parallelogram *CF*
as the base *BC* is to the base *CD*. [V. 11.

Wherefore, *triangles* &c. q.e.d.

Corollary. From this it is plain that triangles and parallelograms which have equal altitudes, are to one an- other as their bases.

For, let the figures be placed so as to have their bases
in the same straight line, and to be on the same side of it ;
and having drawn perpendiculars from the vertices of the
triangles to the bases, the straight line which joins the ver-
tices is parallel to that in which their bases are ; [I. 33.

because the perpendiculars are both equal and parallel to
one another. [I. 28.

Then, if the same construction be made as in the pro- position, the demonstration will be the same.

*THEOREM*.

*If a straight line he drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those sides produced, proportionally ; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section, shall be parallel to the re-maining side of the triangle.* Let *DE* be drawn parallel to *BC*, one of the sides of the triangle *ABC*: *BD* shall be to *DA* as *CE* is to *EA*.

Join *BE*, *CD*.

Then the triangle *BDE* is equal to the triangle *CDE*, because they are on the same base *DE* and between the same parallels *DE*, *BC*. [I. 37.

And *ADE* is another triangle;

and equal magnitudes have the same ratio to the same magnitude; [V. 7.

therefore the triangle *BDE* is to the triangle *ADE* as the triangle *CDE* is to the triangle *ADE*.

But the triangle *BDE* is to the triangle *ADE* as *BD* is to *DA*;

because the triangles have the same altitude, namely, the perpendicular drawn from *E* to *AB*, and therefore they are to one another as their bases. [VI. 1,

For the same reason the triangle *CDE* is to the triangle *ADE* as *CE' is to *EA* *
Therefore

*BD*is to

*DA*as

*CE*is to

*EA'. [V. 11.*

Next, let

*BD*be to

*DA*as

*CE*is to

*EA*, and join

*DE*:

*DE*shall be parallel to

*BC*.

For, the same construction being made, because *BD* is to *DA* as *CE* is to *EA*, [*Hypothesis*.

and as *BD* is to *DA*, so is the triangle *BDE* to the triangle *ADE*, [VI. 1.

and as *CE* is to *EA* so is the triangle *CDE* to the triangle *ADE*; [VI. 1.

therefore the triangle *BDE* is to the triangle *ADE* as the triangle *CDE* is to the triangle *ADE*; [V. 11.

that is, the triangles *BDE* and *CDE* have the same ratio to the triangle *ADE*,

Therefore the triangle *BDE* is equal to the triangle *CDE*. [V. 9.

And these triangles are on the same base *DE* and on the same side of it;

but equal triangles on the same base, and on the same side of it, are between the same parallels; [I. 39.

therefore *DE* is parallel to *BC*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*If the vertical angle of a triangle he bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section shall bisect the vertical angle.*

Let *ABC* be a triangle, and let the angle *BAC* be bisected by the straight line *AD*, which meets the base at *D*: *BD* shall be to *DC* as *BA* is to *AC*.

Through *C* draw *CE* parallel to *DA*, [I. 31.

and let *BA* produced meet *CE* at *E*.

Then, because the straight line *AC* meets the parallels *AD*, *EC*, the angle *ACE* is equal to the alternate angle *CAD*; [I. 29.

but the angle *CAD* is, by hypothesis, equal to the angle *BAD*;

therefore the angle *BAD* is equal to the angle *ACE*. [*Ax*. 1.

Again, because the straight line *BAE* meets the parallels *AD*, *EC*, the exterior angle *BAD* is equal to the interior and opposite angle AEC; [I.29.

but the angle *BAD* has been *shewn* equal to the angle *ACE*;

therefore the angle *ACE* is equal to the angle *AEC*; [*Axiom* 1.

and therefore *AC* is equal to *AE*. [I. 6.

And, because *AD* is parallel to *EC*, [*Constr*.

one of the sides of the triangle *BCE*,

therefore *BD* is to *DC* as *BA* is to *AE*; [VI. 2.

but *AE* is equal to *AC*;

therefore *BD* is to *DC* as *BA* is to *AC*. [V.7.

Next, let *BD* be to *DC* as *BA* is to *AC*, and join *AD*: the angle *BAC* shall be bisected by the straight line *AD*,

For, let the same construction be made. Then *BD* is to *DC* as *BA' 'is to *AC*; [*Hypothesis*. *
and

*BD*is to

*DC*as

*BA*is to

*AE*, [VI. 2.

because

*AD*is, parallel to

*EC*; [

*Construction*.

therefore

*BA*is to

*AC*as

*BA*is to

*AE*; [V. 11.

therefore

*AC*is equal to

*AE*; [V. 9.

and therefore the angle

*AEC*is equal to the angle

*ACE*. [1.5.

But the angle

*AEC*is equal to the exterior angle

*BAD*; [1. 29.

and the angle

*ACE*is equal to the alternate angle

*CAD*; [1. 29.

therefore the angle

*BAD*is equal to the angle CAD;[

*Ax*.l.

that is, the angle

*BAC*is, bisected by the straight line

*AD*.

*if the vertical angle*&c. q.e.d.

*A*.

*THEOREM.*

*If the exterior angle of a triangle, made by producing one of its sides, be bisected by a straight line which also cuts the base produced, the segments between the dividing straight line and the extremities of the base shall have the same ratio which the other sides of the triangle have to one another; and if the segments of the base produced have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section shall bisect the exterior angle of the triangle.*

Let *ABC* be a triangle, and let one of its sides *BA* be produced to *E*; and let the exterior angle *CAE* be bisected by the straight line *AD* which meets the base produced at *D*: *BD* shall be to *DC* as *BA* is to *AC*.

Through *C* draw *CF* parallel to *AD*, [I. 31.

meeting *AB* at *F*.

Then, because the straight line *AC* meets the parallels *AD*, *FC*, the angle *ACF* is equal to the alternate angle *CAD*; [1.29.

but the angle *CAD* is, by hypothesis, equal to the angle *DAE*;

therefore the angle *DAE* is equal to the angle *ACF*. [*Ax*. 1.

Again, because the straight line *FAE* meets the parallels *AD*, *FC*, the exterior angle *DAE* is equal to the interior and opposite angle *AFC*; [I. 29.

but the angle *DAE* has been shewn equal to the angle *ACF*;

therefore the angle *ACF' 'is, equal to the angle *AFC*; [*Ax*. 1. *
and therefore

*AC*is equal to

*AF*. [I. 6.

And, because

*AD*is paralled to

*FC*[

*Construction*.

one of the sides of the triangle

*BCF*;

therefore

*BD*is to

*DC*as

*BA*is to

*AF*; [VI. 2.

but

*AF*is equal to

*AC*;

therefore

*BD*is to

*DC*as

*BA*is to

*AC*[V. 7.

Next, let

*BD*be to

*DC*as

*BA*is to

*AC*; and join

*AD*: the exterior angle

*CAE*shall be bisected by the straight line

*AD*.

For, let the same construction be made.

Then *BD* is to *DC* as *BA* is to *AC*; [*Hypothesis*.

and *BD* is to *DC* as *BA* is to *AF*; [VI. 2.

therefore *BA* is to *DC* as *BA* is to *AF*; [V. 11.

therefore *AC* is equal to *AF* [V. 9.

and therefore the angle *ACF* is equal to the angle *AFC*. [1. 5.

But the angle *AFC* is equal to the exterior angle *DAE*; [1. 29.

and the angle *ACF* is equal to the alternate angle *CAD*; [1. 29.

therefore the angle *CAD* is equal to the angle *DAE*; [*Ax*. 1.

that is, the angle *CAE* is bisected by the straight line *AD*.

Wherefore, *if the exterior angle* &c. q.e.d.

*THEOREM*.

*The sides about the equal angles of triangles which are equiangular to one another are proportionals; and those which are opposite to the equal angles are homologous sides that is, are the antecedents or the consequents of the ratios.*

Let the triangle *ABC* be equiangular to the triangle *DCE*, having the angle *ABC* equal to the angle *DCE*, and the angle *ACB* equal to the angle *DEC*, and consequently the angle *BAC* equal to the angle *CDE* the sides about the equal angles of the triangles *ABC*, *DCE*, shall be proportionals; and those shall be the homologous sides, which are opposite to the equal angles.

Let the triangle *DCE* be placed so that its side *CE* may be contiguous to *BC*, and in the same straight line with it. [I. 22.

Then the angle *BCA* is equal to the angle *CED* ; [*Hyp*.

add to each the angle *ABC* ;

therefore the two angles *ABC*, *BCA* are equal to the two angles *ABC*, *CED* ; [*Axiom* 2.

but the angles *ABC*, *BCA* are together less than two
right angles; [1.17.

therefore the angles ABC, CED are together less than
two right angles ;

therefore *BA* and *ED*, if produced, will meet. [*Axiom* 12.

Let them be produced and meet at the point *F*.

Then, because the angle *ABC* is equal to the angle
DCE, [*Hypothesis*.

*BF* is parallel to *CD* ; [*I. 28. *
and because the angle

*ACB*Is equal to the angle

*DEC*, [

*Hyp*.

*AC*is parallel to

*FE*. [I. 28.

Therefore

*FACD*is a parallelogram;

and therefore

*AF*is equal to

*CD*, and

*AC*is equal to

*FD*. [1. 34.

And, because *AC* be parallel to *FE*, one of the sides of
the triangle *FBE*,

therefore *BA* is to *AF* as *BC* is to *CE* ; [VI. 2.

but *AF* is equal to *CD* ;

therefore *BA* is to *CD* as *BC* is to *CE* ; [V. 7.

and, alternately, *AB* is to *BC* as *DC* is to *CE*, [V. 16.

Again, because *CD* is parallel to *BF*,

therefore *BC* is to *CE* as *FD* is to *DE* ; [VI. 2.

but *FD* is equal to *AC*;

therefore *BC* is to *CE* as *AC* to *DE* ; [V. 7.

and, alternately, *BC* is to *CA* as *CE* is to *ED*. [V. 16.

Then, because it has been shewn that *AB* is to *BC* as *DC*
is to *CE*, and that *DC* is to *CA* as *CE* is to *ED* ;

therefore, ex aequali, *BA* is to *AC* as *CD* is to *DE*. [V. 22.

Wherefore, *the sides* &c. q.e.d.

*THEOREM*.

*If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular to one another, and shall have those angles equal which are opposite to the homologous sides.* Let the triangles *ABC*, *DEF* have their sides proportional, so that *AB* is to *BC* as *DE* is to *EF*; and *BC* to *CA* as *EF* is to *FD*; and, consequently, ex aequali, *BA* to *AC* as *ED* is to *DF*: the triangle *ABC* shall be equian^ilar to the triangle *DEF*, and they shall have those angles equal which are opposite to the homologous sides, namely, the angle *ABC* equal to the angle *DEF*, and the angle *BCA* equal to the angle *EFD*, and the angle *BAC* equal to the angle *EDF*.

At the point *E*, in the straight line *EF*, make the angle *FEG* equal to the angle *ABC*; and at the point *F*, in the straight line *EF*, make the angle *EFG* equal to the angle *BCA*; [I. 23.

therefore the remaining angle *EGF* is equal to the remaining angle *BAC*.

Therefore the triangle *ABC* is equiangular to the triangle *GEF*;

and therefore they have their sides opposite to the equal angles proportionals; [VI. 4.

therefore *AB* is to *BC* as *GE* is to *EF*.

But *AB* is to *BC* as *DE* is to *EF*: [*Hypothesis*.

therefore *DE* is to *EF* as *GE* is to *EF*; [V. 11.

therefore *DE* is equal to *GF*. [V. 9.

For the same reason, *DF* is equal to *GF*.

Then, because in the two triangles *DEF*, *GEF*, *DE* is equal to *GE*, and *EF* is common;

the two sides *DE*, *EF* are equal to the two sides *GE*, *EF*, each to each;

and the base *DF* is equal to the base *GF*;

therefore the angle *DEF* is equal to the angle *GEF*, [I. 8.

and the other angles to the other angles, each to each, to which the equal sides are opposite. [I. 4,

therefore the angle *DFE* is equal to the angle *GFE*, and the angle *EDF* is equal to the angle *EGF*. And, because the angle *DEF* is equal to the angle *GEF*,
and the angle *GEF* is equal to the angle *ABC*, [*Constr*.

therefore the angle *ABC* is equal to the angle *DEF*. [*Ax*. 1.

For the same reason, the angle *ACB* is equal to the angle
*DFE*, and the angle at *A* is equal to the angle at *D'. *
Therefore the triangle

*ABC*is equiangular to the triangle

*DEF*.

Wherefore, *if the sides* &c. q.e.d.

*THEOREM*.

*If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles another, and shall have those angles equal which are opposite to the homologous sides.*

Let the triangles *ABC*, *DEF* have the angle *BAC* in
the one, equal to the angle *EDF* in the other, and the
sides about those angles proportionals, namely, *BA* to *AC*
as *ED* is to *DF*: the triangle *ABC* shall be equiangular to
the triangle *DEF*, and shall have the angle *ABC* equal to
the angle *DEF*, and the angle *ACB* equal to the angle *DFE*.

At the point *D*, in the
straight line *DF*, make the
angle *FDG* equal to either
of the angles *BAC*, *EDF* ;
and at the point *F*, in the
straight line *DF*, make
the angle *DFG* equal to
the angle *ACB*; [I. 23.

therefore the remaining angle at *G* is equal to the remain-
ing angle at *B*.

Therefore the triangle *ABC* is equiangular to the triangle
*DGF*;

therefore *BA* is to *AC* as *GD* is to *DF*. [VI. 4.

But *BA* is to *AC* as *ED* is to *DF*; [*Hypothesis*.

therefore *ED* is to *DF* as *GD* is to *DF*; [V. 11.

therefore *ED* is equal to *GD*. [V. 9.

And *DF* is common to the two triangles *EDF*, *GDF* therefore the two sides *ED*,*DF* are equal to the two sides *GD*, *DF* each to each;

and the angle *EDF* is equal to the angle *GDF*; [*Constr*. therefore the base *EF* is equal to the base *GF*, and the triangle *EDF* to the triangle *GDF*, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; [I. 4.

therefore the angle *DFG* is equal to the angle *DFE*, and the angle at *G* is equal to the angle at *E*.

But the angle *DFG* is equal to the angle *ACB*; [*Constr*.

therefore the angle *ACB* is equal to the angle *DFE*. [*Ax*. 1.

And the angle *BAC* is, equal to the angle *EDF*; [*Hypothesis*.

therefore the remaining angle at *B* is equal to the remaining angle at *E*.

Therefore the triangle *ABC* is equiangular to the triangle *DEF*.

Wherefore, *if two triangles* &c. q.e.d.

*THEOREM*.

*If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; then if each of the remaining angles be either less, or not less, than a right angle, or if one of them he a right angle, the triangles shall he equiangular to one another, and shall have those angles equal about touch the sides are proportionals.*

Let the triangles *ABC*, *DEF* have one angle of the one equal to one angle of the other, namely, the angle *BAC* equal to the angle *EDF*, and the sides about two other angles *ABC*, *DEF*, proportionals, so that *AB* is to *BC* as *DE* is to *EF*; and, first, let each of the remaining angles at *C* and *F* be less than a right angle: the triangle *ABC* shall be equiangular to the triangle *DEF* and shall have the angle *ABC* equal to the angle *DEF*, and the angle at *C* equal to the angle at *F*.

For, if the angles *ABC*, *DEF* be not equal, one of them must be greater than the other.

Let *ABC* be the greater, and at the point *B*, in the straight line *AB*, make the angle *ABC* equal to the angle *DEF*. [I, 23.

Then, because the angle at *A* is equal to the angle at *D*, [*Hyp*.

and the angle *ABG* is equal to the angle *DEF*, [*Constr*.

therefore the remaining angle *AGB* is equal to the remaining angle *DFE*;

therefore the triangle *ABG* is equiangular to the triangle *DEF*.

Therefore *AB* is to *BG* as *DE* is to *EF*. [VI. 4.

But *AB* is to *BG* as *DE* is to *EF*; [*Hypothesis*.

therefore *AB* is to *BC* as *AB* is to *BG*; [V. 11.

therefore *BC* is equal to *BG*; [V. 9.

and therefore the angle *BCG* is equal to the angle *BGC*. [I. 5.

But the angle *BCG* is less than a right angle; [*Hyp*.

therefore the angle *BGC* is less than a right angle;

and therefore the adjacent angle *AGB* must be greater than a right angle. [I. 13.

But the angle *AGB* was shewn to be equal to the angle at *F*;

therefore the angle at *F* is greater than a right angle.

But the angle at *F* is less than a right angle; [*Hypothesis*.

which is absurd.

Therefore the angles *ABC* and *DEF* are not unequal; that is, they are equal.

And the angle at *A* is equal to the angle at *D*; [*Hypothesis*.

therefore the remaining angle at *C* is equal to the remaining angle at *F*;

therefore the triangle *ABC* is equiangular to the triangle *DEF*. Next, let each of the angles at *C* and *F* be not less than a right angle: the triangle *ABC* shall be equiangular to the triangle *DEF*.

For, the same construction being made, it may be shewn in the same manner, that *BC* is equal to *BG*;

therefore the angle *BCG* is equal to the angle *BGC*. [I. 5.

But the angle *BCG* is not less than a right angle; [*Hyp*.

therefore the angle *BCG* is not less than a right angle; that is, two angles of the triangle *BCG* are together not less than two right angles; which is impossible. [I, 17.

Therefore the triangle *ABC* may be shewn to be equiangular to the triangle *DEF*, as in the first case.

Lastly, let one of the angles at *C* and *F* be a right angle, namely, the angle at *C*: the triangle *ABC* shall be equiangular to the triangle *DEF*.

For, if the triangle *ABC* be not equiangular to the triangle *DEF*, at the point *B*, in the straight line *AB*, make the angle *ABG* equal to the angle *DEF*. [I. 23.

Then it may be shewn, as in the first case, that *BC* is equal to *BG*; therefore the angle *BCG* is equal to the angle *BGC* [I. 5.

But the angle *BCG* is a right angle: [*Hypothesis*.

therefore the angle *BGC* is a right angle;

that is, two angles of the triangle *BCG* are together equal to two right angles; which is impossible. [I. 17.

Therefore the triangle *ABC* is equiangular to the triangle *DEF*.

*if two triangles*&c. q.e.d.

*THEOREM*.

*In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.*

Let *ABC* be a right-angled triangle, having the right angle *BAC*; and from the point *A*, let *AD* be drawn perpendicular to the base *BC*: the triangles *DBA*, *DAC* shall be similar to the whole triangle *ABC*, and to one another.

For, the angle *BAC* is equal to the angle *BDA*, each of them being a right angle, [*Axiom* 11. and the angle at *B* is common to the two triangles *ABC*, *DBA*;

therefore the remaining angle *ACB* is equal to the remaining angle *DAB*.

Therefore the triangle *ABC* is equiangular to the triangle *DBA*, and the sides about their equal angles are proportionals; [VI. 4.

therefore the triangles are similar. [VI. *Definition* 1.

In the same manner it may be shewn that the triangle *DAC* is similar to the triangle *ABC*.

And the triangles *DBA*, *DAC* being both similar to the triangle *ABC*, are similar to each other.

Wherefore, *in a right-angled triangle* &c. q.e.d.

Corollary. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base and the segment of the base adjacent to that side.

For, in the triangles *DBA*, *DAC*,

*BD* is to *DA* as *DA* is to *DC*; [VI. 4.

and in the triangles *ABC*, *DBA*,

*BC* is to *BA* as *BA* is to *BD*; [VI. 4.

and in the triangles *ABC*,*DAC*,

*BC*is to

*CA*as

*CA*is to

*CD*. [VI. 4.

*PROBLEM*.

*From a given straight line to cut off any part required.*

Let *AB* be the given straight line: it is required to cut off any part from it.

From the point *A* draw a straight line *AG*, making any angle with *AB*;

in *AG* take any point *D*, and take *AC* the same multiple of *AD*, that *AB* is of the part which is to be cut off from it; join *BC*, and draw *DE* parallel to it. *AE* shall be the part required to be cut off.

For, because *ED* is parallel to *BC*, [*Construction*.

one of the sides of the triangle *ABC*,

therefore *CD* is to *DA* as *BE* is to *EA*; [VI. 2.

and, by composition, *CA* is to *AD* as *BA* is to *AE*. [V. 18.

But *CA* is a multiple of *AD*; [*Construction*.

therefore *BA* is the same multiple of *AE*; [V. *D*.

that is, whatever part *AD* is of *AC*, *AE* is the same part of *AB*.

Wherefore, *from the given straight line AB, the part required has been cut off*. q.e.f.

*PROBLEM.*

*To divide a given straight line similarly to a given divided straight line, that is, into parts which shall have the same ratios to one another, that the parts of the given divided straight line have.*

Let *AB* be the straight line given to be divided, and *AC* the given divided straight line: it is required to divide *AB* similarly to *AC*.

Let *AC* be divided at the points *D*, *E* and let *AB*, *AC* be placed so as to contain any angle, and join *BC*; through the point *D*, draw *DF* parallel to BC, and through the point E draw EG parallel to BC. [I. 31.

*AB* shall be divided at the points *F*, *G*, similarly to *AC*. Through *D* draw *DHK* parallel to *AB*. [I. 31.

Then each of the figures *FH*,*HB* is a parallelogram; therefore *BH* is equal to *FG*, and *HK* is equal to *GB*. [1.34.

Then, because *HF* is parallel to *KC*, [*Construction*.

one of the sides of the triangle *DKC*,

therefore *KH* is to *HD* as *CE* is to *ED*. [VI. 2.

But *KH* is equal to *BG*, and *HD is equal to *GF*; therefore *BG* is to *GF* as *GE* is to *ED*. [V. 7.*

Again, because *FD* is parallel to *GE*, [*Construction*.

one of the sides of the triangle *AGE*, therefore *GF* is to *FA* as *ED* is to *DA*. [VI. 2.

And it has been shewn that *BG* is to *GF* as *CE* is to *EB*. Therefore *BG* is to *GF* as *CE* is to *ED*, and *GF* is to *FA* as *ED' 'is to *BA*.*

Wherefore *the given straight line AB is divided similarly to the given divided straight line AC*. q.e.f.

*PROBLEM*.

*To find a third proportional to two given straight lines.*

Let *AB*, *AC* be the two given straight lines: it is required to find a third proportional to *AB*, *AC*.

Let *AB*, *AC* be placed so as to contain any angle; produce *AB*, *AC*, to the points *D*, *E*; and make *BD* equal to *AC*; [I. 3.

join *BC*, and through *D draw *DE* parallel to *BC*. [I. 31.*

*CE shall be a third proportional to*AB

*,*AC

*.*

For, because

*BC' 'is parallel to*BE

*, [*Construction

*.*

one of the sides of the triangle

*ABE*, therefore

*AB*is to

*BD*as

*AC*is to

*CE*; [VI. 2.

but

*BD*is equal to

*AC*; [

*Construction*.

therefore

*AB*is to

*AC*as

*AC*is to

*CE*. [V. 7.

*to the two given straight lines AB, AC, a third proportional CE is found*. q.e.f.

*PROBLEM*.

*To find a fourth proportional to three given straight lines.*

Let *A*, *B*, *C* be the three given straight lines: it is required to find a fourth proportional to *A*, *B',' *C*.*

Take two straight lines, *DE*, *DF* containing any angle *EDF*; and in these make *DG* equal to *A*, *GE* equal to *B*, and *DH* equal to *C*; [I. 3.

join *GH*, and through *E* draw *EF* parallel to *GH*. [I. 31.

*HF* shall be a fourth propertional to *A*,*B*,*C*

For, because *GH* is parallel to *EF*, [*Construction*.

one of the sides of the triangle *DEF*, therefore *DG* is to *GE* as *DH* is to *HF*. [VI. 2.

But *DG* is equal to *A*, *GE* is equal to *B*, and *DH* is equal to *C*; [*Construction*.

therefore *A* is to *B* as *C* is to *HF*. [V. 7.

Wherefore to the three given straight lines *A*, *B*, *C*, a fourth proportional *HF* is found, q.e.f.

*PROBLEM*.

*To find a mean proportional between two given straight lines*.

Let *AB*, *BC* be the two given straight lines: it is required to find a mean proportional between them.

Place *AB*,*BC* in a straight line, and on *AC* describe the semicircle *ADC*; from the point *B* draw *BD* at right angles to *AC*. [I. 11.

*BD* shall be a mean proportional between *AB* and *BC*.

Join *AD*, *DC*.

Then, the angle *ADC*, being in a semicircle, is a right angle; [III. 31.

and because in the right-angled triangle *ADC*, *DB* is drawn from the right angle perpendicular to the base,

therefore *DB* is a mean proportional between *AB*, *BC*, the segments of the base. [VI. 8, *Corollary*.

Wherefore, *between the two given straight lines. AB*, *BC, a mean proportional *DB* is found*, q.e.f.

*THEOREM*.

*Equal parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional; and parallelograms which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another*.

Let *AB*, *BC* be equal parallelograms, which have the angle *FBD* equal to the angle *EBG*: the sides of the parallelograms about the equal angles shall be reciprocally proportional, that is, *DB* shall be to *BE* as *GB* is to *BF*.

Let the parallelograms be placed, so that the sides *DB*, *BE* may be in the same straight line;

therefore also *FB*, *BG* are in one straight line. [I. 14.

Complete the parallelogram *FE*.

Then, because the parallelogram *AB* is equal to the parallelogram *BC*, [*Hypothesis*.

and that *FE* is another parallelogram,

therefore *AB* is to *FE* as *BC* is to *FE*. [V. 7.

But *AB* is to *FE* as the base *DB* is to the base *BE*, [VI. 1.

and *BC* is to *FE* as the base *GB* is to the base *BF*; [VI. 1.

therefore *DB* is to *BE* as *GB* is to *BF*. [V. 11.

Next, let the angle *FBD* be equal to the angle *EBG*, and let the sides about the equal angles be reciprocally proportional, namely, *DB* to *BE* as *GB* is to *BF*: the parallelogram *AB* shall be equal to the parallelogram *BC*.

For, let the same construction be made.

Then, because *DB* is to *BE* as *GB* is to *BF*, [*Hypothesis*

and that *DB* is to *BE* as the parallelogram *AB* is to the parallelogram *FE*, [VI. 1.

and that *GB* is to 'BF* as the parallelogram *BG* is to the parallelogram 'FE'; [VI. 1.*
therefore the parallelogram

*AB*is to the parallelogram

*FE*as the parallelogram

*BC*is to the parallelogram

*FE*; [V. 11.

therefore the parallelogram

*AB*is equal to the parallelogram

*BC*. [V. 9.

Wherefore, *equal parallelograms* &c. q.e.d.

*THEOREM*.

*Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional; and triangles which have one angle of the one equal to one angle of the other, and their sides about the equM angles reciprocally proportional, are equal to one another.*

Let *ABC*, *ADE* be equal triangles, which have the angle *BAC* equal to the angle *DAE*: the sides of the triangles about the equal angles shall be reciprocally proportional; that is, *CA* shall be to *AD* as *EA* is to *AB*.

Let the triangles be placed so that the sides *CA*, *AD* may be in the same straight line,

therefore also *EA*, *AB* are in one straight line; [1. 14.

join *BD*.

Then, because the triangle *ABC* is equal to the trian-
gle *ADE*, [*Hypothesis*.

and that *ABD* is another triangle,

therefore the triangle *ABC* is to the triangle *ABD* as the
triangle *ADE* is to the triangle *ABD*. [V. 7.

But the triangle *ABC* is to the triangle *ABD* as the base
*CA* is to the base *AD*, [VI. 1.

and the triangle *ADE* is to the triangle *ABD* as the base
*EA* is to the base *AB* ; [VI. 1.

therefore *CA* is to *AD* as *EA* is to *AB*. [V. 11.

Next, let the angle *BAC* be equal to the angle *DAE*,
and let the sides about the equal angles be reciprocally
proportional, namely, *CA* to *AD* as *EA* is to *AB*: the
triangle *ABC* shall be equal to the triangle *ADE*.

For, let the same construction be made.
Then, because *CA* is to *AD' 'as *EA' 'is to *AB*, [*Hypothesis*.

and that *CA* is to *AD* as the triangle *ABC* is to the
triangle *ABD*, [VI. 1.

and that *EA* is to *AB* as the triangle *ADE* is to the
triangle *ABD*, [VI. 1.

therefore the triangle *ABC* is to the triangle *ABD* as the
triangle *ADE* is to the triangle *ABD* ; [V. 11.

therefore the triangle *ABC* is equal to the triangle *ABD* [V. 9.

Wherefore, *equal triangles* &c. q.e.d.

PROPOSITION 16. *THEOREM*.

*If four straight lines he proportionals, the rectangle contained by the extremes is equal to the rectangle con-tained by the means; and if the rectangle contained by the' extremes he equal to the rectangle contained by the means, the four straight lines are proportionals.* Let the four straight lines *AB*, *CD*, *E*, *F*, be proportionals, namely, let *AB* be to *CD* as *E* is to *F*: the rectangle contained by *AB* and *F* shall be equal to the rectangle contained by *CD* and *E*.

From the points *A*, *C*, draw *AG*, *CH* at right angles to *AB*, *CD*; [I. 11.

make *AG* equal to *F*, and *CH* equal to *E*; [I. 3.

and complete the parallelograms *BG*, *DH*. [I.31 .

Then, because *AB* is to *CD* as *E* is to *F*, [*Hyp*.

and that *E* is equal to *CH*, and *F* is equal to *AG*, [*Construction*.

therefore *AB* is to *CD* as *CH* is to *AG*; [V. 7.

that is, the sides of the parallelograms *BG*, *DH* about the equal angles are reciprocally proportional;

therefore the parallelogram *BG* is equal to the parallelogram *DH*. [VI. 14.

But the parallelogram *BG* is contained by the straight lines *AB* and *F*, because *AG* is equal to F, [*Construction*.

and the parallelogram *DH* is contained by the straight lines *CD* and *E*, because *CH* is equal to *E*;

therefore the rectangle contained by *AB* and *F* is equal to the rectangle contained by *CD* and *E*.

Next, let the rectangle contained by *AB* and *F* be equal to the rectangle contained by *CD* and *E*: these four straight lines shall be proportional, namely, *AB* shall be to *CD* as *E* is to *F*.

For, let the same construction be made.

Then, because the rectangle contained by *AB* and *F* is equal to the rectangle contained by *CD* and *E*, [*Hypothesis*.

and that the rectangle *BG* is contained by *AB* and *F*, because *AG* is equal to *F*, [*Construction*.

and that the rectangle *DH* is contained by *CD* and *E*, because *CH* is equal to *E*, [*Construction*.

[*Axiom* 1.therefore the parallelogram *BG* is equal to the parallelogram *DH*.

[VI. 14.And these parallelograms are equiangular to one another; therefore the sides about the equal angles are reciprocally proportional;

therefore *AB* is to *CD* as *CH* is to *AG*.

[*Constr*.But *CH* is equal to *E*, and *AG* is equal to *F*;

[V. 7.therefore *AB* is to *CD* as *E* is to *F*.

Wherefore, *if our straight lines* &c. q.e.d.

PROPOSITION 17. *THEOREM*.

*If three straight lines he proportionals, the rectangle contained by the extremes is equal to the square on the mean; and if the rectangle contained hy the extremes he equal to the square on the mean, the three straight lines are proportionals.*

Let the three straight lines *A*, *B*, *C* be proportionals, namely, let *A* be to *B* as *B* is to *C*: the rectangle contained by *A* and *C* shall be equal to the square on *B*.

Take *D* equal to *B*.

[*Hyp.*Then, because *A* is to *B* as *B* is to *C*,

and that *B* is equal to *D*,

[V. 7.therefore *A* is to *B* as *D* is to *C*.

[VI. 16.But if four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means;

therefore the rectangle contained by *A* and *C* is equal to the rectangle contained by *B* and *D*.

[*Construction*.But the rectangle contained by *B* and *D* is the square on *B* because *B* is equal to *D*;

therefore the rectangle contained by *A* and *C* is equal to the square on *B*.

Next, let the rectangle contained by *A* and *C* be equal to the square on *B*: *A* shall be to *B* as *B is to *C*,*

For, let the same construction be made.

Then, because the rectangle contained by *A* and *C* is equal to the square on *B*, [*Hypothesis*.

and that the square on *B* is equal to the rectangle contained by *B* and *D*, because *B* is equal to *D*, [*Construction*.

therefore the rectangle contained by *A* and *C* is equal to the rectangle contained by *B* and *D*.

But if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals; [VI. 16.

therefore *A* is to *D* as *B* is to *C*.

But *B* is equal to *D*; [*Construction*.

Therefore *A* is to *B* as *B* is to *C*. [V. 7.

Wherefore, *if three straight lines* &c. q.e.d.

PROPOSITION 18. *PROBLEM*.

*On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure*.

Let *AB* be the given straight line, and *CDEF* the given rectilineal figure of four sides: it is required to describe on the given straight line *AB*, a, rectilineal figure, similar and similarly situated to *CDEF*.

Join *DF* at the point *A*, in the straight line *AB*, make the angle *BAC* equal to the angle; and at the point *B', in the straight line *AB*, make the angle *ABG* equal to the angle*CDF*;[I.23. therefore the remaining angle *AGB

*is equal to the remain-*ing angle

*CFD*,

and the triangle

*AGB*is equiangular to the triangle

*CFD*,

Again, at the point

*B*, in the straight line

*BG*, make the angle

*GBH*equal to the angle

*FDE*; and at the point

*G*, in the straight line

*BG*, make the angle

*BGH*equal to the angle

*DFE*; [1.23.

therefore the remaining angle

*BHG*is equal to the re- maining angle

*DEF*,

and the triangle

*BHG*is equiangular to the triangle

*DEF*.

Then, because the angle *AGB* is equal to the angle *CFD*,
and the angle *BGH* equal to the angle *DFE*; [*Construction*.

therefore the whole angle *AGH* is equal to the whole
angle *CFE*. [*Axiom 2. *
For the same reason the angle

*ABH*is equal to the angle

*CDE*.

And the angle

*BAG*is equal to the angle

*DCF*, and the angle

*BHG*is equal to the angle

*DEF*.

Therefore the rectilineal figure

*ABHG*is equiangular to the rectilineal figure

*CDEF*.

Also these figures have their sides about the equal angles proportionals.

For, because the triangle *BAC* is equiangular to the triangle
*DCF*, therefore *BA* to *AC* as *DC* is to *CF*. [VI. 4.

And, for the same reason, *AC* is to *GB* as *CF* is to *FD*,
and *BC* is to *GH* as *DF* is to *FE* ;

therefore, ex aequali, *AG* is to *GH* as *CF* is to *FB*. [V. 22.

In the same manner it may be shewn that *AB* is to *BH*
as *CD* is to *DE*.

And *GH* is to *HB* as *FE* is to *ED*. [VI. 4.

Therefore, the rectilineal figures *ABHG* and *CDEF*
are equiangular to one another, and have their sides about
the equal angles proportionals ;
therefore they are similar to one another. [VI. *Definition* 1.

Next, let it be required to describe on the given straight
line *AB*, a rectilineal figure, similar, and similarly situated,
to the rectilineal figure *CDKEF* of five sides.
Join DE, and on the given straight line *AB* describe, as in the former case, the rectilineal figure *ABHG*, similar, and similarly situated to the rectilneal figure *CDEF* of four sides. At the point *B' in the straight line *BH*, make the angle *HBL* equal to the angle *EDK* and at the point *H*, in the straight line *BH* make the angle *BHL* equal to the angle *DEK*; [I. 23.*
therefore the remaining angle at

*L*is equal to the remaining angle at

*K*.

Then, because the figures *ABHG*, *CDEF* are similar, the angle *ABH* is equal to the angle *CDE*; [VI. *Def*. 1.

and the angle *HBL* is equal to the angle *EDK* [*Constr*.

therefore the whole angle *ABL* is equal to the whole angle *CDK*, [*Axiom* 2.

For the same reason the whole angle *GHL* is equal to the whole angle *FEK*.

Therefore the five-sided figures *ABHG* and *CDKEF* are equiangular to one another.

And, because the figures *ABHG* and *CDEF* are similar, therefore *AB* is to *BH* as *CD* is to *DE* [VI. *Definition* 1.

but *BH* is to *BL* as *DE* is to DK; [VI. 4.

therefore, ex aequali, *AB* is to *BL* as *CD is to *DK*. [V. 22. *
For the same reason,

*GH*is to

*HL*as

*FE*is to

*EK*.

And

*BL*is to

*LH*as

*DK*is to

*KE*. [VI. 4.

Therefore, the five-sided figures *ABLHG* and *CDKEF* are equiangular to one another, and have their sides about the equal angles proportionals;

therefore they are similar to one another. [VI. *Definition* 1.

PROPOSITION 19. *THEOREM*.

*Similar triangles are to one another in the duplicate ratio of their homologous sides.*

Let *ABC* and *DEF* be similar triangles, having the angle *B* equal to the angle *E*, and let *AB* be to *BC* as *DE* is to *EF* so that the side *BC* is homologous to the side *EF*: the triangle *ABC* shall be to the triangle *DEF* in the duplicate ratio of *BC* to *EF*.

Take *BG* a third proportional to *BG* and *EF*, so that *BG* may be to *EF* as *EF* is to *BG*; [VI. 11.

and join *AG*.

Then, because *AB* is to BC as *DE' is to *EF*, [*Hypothesis*. *
therefore, alternately,

*AB*is to

*DE*as

*C*is to

*EF*; [V. 16.

but

*BC*is to

*EF*as

*EF*is to

*BG*; [

*Construction*.

therefore

*AB*is to

*DE*as

*EF*is to

*BG*; [V. 11.

that is, the sides of the triangles

*ABG*and

*DEF*, about their equal angles, are reciprocally proportional;

but triangles which have their sides about two equal angles reciprocally proportional are equal to one another, [VI. 15.

therefore the triangle

*ABG*is equal to the triangle

*DEF*.

And, because *BG* is to *EF* as *EF* is to BG,

therefore *BG* has to *BG* the duplicate ratio of that which *BC* has to *EF*. [V. *Definition* 10.

But the triangle *ABC* is to the triangle *ABG* as *BC* is to *BG*; [VI. 1.

therefore the triangle *ABC* has to the triangle *ABG* the duplicate ratio of that which *BC* has to *EF*. But the triaingle *ABG* was shewn equal to the triangle *DEF*;

therefore the triangle *ABC* has to the triangle *DEF* the duplicate ratio of that which *BC* has to *EF*. [V. 7.

Wherefore, *similar triangles* &c. q.e.d.

Corollary. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle described on the first to a similar and similarly described triangle on the second.

PROPOSITION 20. *THEOREM*.

*Similar polygons may he divided into the same number of similar triangles, having the same ratio to one another that the polygons have; and the polygons are to one another in the duplicate ratio of their homologous sides.*

Let *ABCDE*, *FGHKL* be similar polygons, and let *AB* be the side homologous to the side *FG*: the polygons *ABCDE*, *FGHKL* may be divided into the same number of similar triangles, of which each shall have to each the same ratio which the polygons have; and the polygon *ABCDE* shall be to the polygon *FGHKL* in the duplicate ratio of *AB* to *FG*.

Join *BE*, *EC*, *GL*, *LH*.

Then, because the polygon *ABCDE* is similar to the polygon *FGHKL*, [*Hypothesis*.

the angle *BAE* is equal to the angle *GFL*, and *BA* is to *AE* as *GF* is to *FL*. [VI. *Definition* 1.

And, because the triangles *ABE* and *FGL* have one angle of the one equal to one angle of the other, and the sides
about these equal angles proportionals,

therefore the triangle ABE is equiangular to the triangle *FGL*, [VI. 6.

and therefore these triangles are similar; [VI. 4.

therefore the angle *ABE* is equal to the angle *FGL*.

But, because the polygons are similar, [*Hypothesis*.

therefore the whole angle *ABC* is equal to the whole angle
*FGH*, [VI. *Definition* 1.

therefore the remaining angle *EBC* is equal to the remain-
ing angle *LGH*. [*Axiom* 3.

And, because the triangles *ABE* and *FGL* are similar,

therefore *EB* is to *BA* as *LG* is to *GF*;

and also, because the polygons are similar, [*Hypothesis*.

therefore *AB* is to *BC* as *FG* is to *GH* ; [VI. *Definition* 1.

therefore, ex aequali, *EB* is to *BC* as *LG* is to *GH*; [V. 22.

that is, the sides about the equal angles *EBC* and *LGH*
are proportionals ;

therefore the triangle *EBC* is equiangular to the triangle
*LGH*; [VI. 6.

and therefore these triangles are similar. [VI. 4.

For the same reason the triangle *ECD* is similar to the
triangle *LHK*.

Therefore the similar polygons *ABCDE*, *FGHKL* may be
divided into the same number of similar triangles.

Also these triangles shall have, each to each, the same
ratio which the polygons have, the antecedents being *ABE*,
*EBC*, *ECD*, and the consequents *FGL*, *LGH*, *LHK*; and
the polygon *ABCDE* shall be to the polygon *FGHKL* in
the duplicate ratio of *AB* to *FG*.

For, because the triangle *ABE* is similar to the tri-
angle *FGL*,

therefore *ABE* is to *FGL* in the duplicate ratio of *EB*
to *LG*. [VI. 19.

For the same reason the triangle *EBC* is to the triangle
*LGH* in the duplicate ratio of *EB* to *LG*.

Therefore the triangle *ABE* is to the triangle *FGL* as the
triangle *EBC* is to the triangle *LGH*. [V. 11.

Again, because the triangle *EBC* is similar to the tri-
angle *LGH*,

therefore *EBC* is to *LGH* in the duplicate ratio of *EC*
to *LH*. [VI. 19.

For the same reason the triangle *ECD* is to the triangle *LHK* in the duplicate ratio of *EC* to *LH*.

Therefore the triangle *EBC* is to the triangle *LGH* as the triangle *ECD* is to the triangle *LHK*. [V. 11.

But it has been shewn that the triangle *EBC is to the triangle *LGH* as the triangle *ABE* is to the triangle *FGL*.*
Therefore as the triangle

*ABE*is to the triangle

*FGL*, so is the triangle

*EBC*to the triangle

*LGH*, and the triangle

*ECD*to the triangle

*LHK*; [V. 11.

and therefore as one of the antecedents is to its consequent so are all the antecedents to all the consequents; [V. 12.

that is, as the triangle

*ABE*is to the triangle

*FGL*so is the polygon

*ABCDE*to the polygon

*FGHKL*.

But the triangle *ABE* is to the triangle *FGL* in the duplicate ratio of the side *AB* to the homologous side *FG*; [VI. 19.

therefore the polygon *ABCDE* is to the polygon *FGHKL* in the duplicate ratio of the side *AB* to the homologous side *FG*.

Wherefore, *similar polygons* &c. q.e.d.

Corollary 1. In like manner it may be shewn that similar four- sided figures, or figures of any number of sides, are to one another in the duplicate ratio of their homologous sides; and it has already been shewn for triangles; therefore universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

Corollary 2. If to *AB* and *FG*, two of the homologous sides, a third proportional *M* be taken, [VI. 11.

then *AB* has to *M* the duplicate ratio of that which *AB* has to *FG*. [V. *Definition* 10.

But any rectilineal figure described on 'AB' is to the similar and similarly described rectilineal figure on *FG* in the duplicate ratio of *AB* to *FG*, [*Corollary* 1.

Therefore as *AB* is to *M*, so is the figure on *AB* to the figure on *FG*; [V. 11.

and this was shewn before for triangles. [VI. 19, *Corollary*.

Wherefore, *universally, if three straight lines be propor-tionals, as the first is to the third, so is any rectilineal figure described on the first to a similar and similarly described rectilineal figure on the second.*

PROPOSITION 21. *THEOREM*.

*Rectilineal figures which are similar to the same rectilineal figure, are also similar to each other.*

Let each of the rectilineal figures *A* and *B* be similar to the rectilineal figure *C*: the figure *A* shall be similar to the figure *B*.
For, because *A* is similar to *C*, [*Hyp.*

*A*is equiangular to

*C', and*A

*and*C

*have their sides about the equal angles proportionals. [VI.*Def

*. 1.*

Again, because

*B*is similar to

*C*, [

*Hyp*.

*B*is equiangular to

*C*, and

*B*and

*C*have their sides about the equal angles proportionals. [VI.

*Definition*1.

Therefore the figures

*A*and

*B*are each of them equiangular to

*C*, and have the sides about the equal angles of each of them and of

*C*proportionals.

Therefore

*A*is equiangular to

*B*, [

*Axiom*1.

and

*A*and

*B*have their sides about the equal angles proportionals; [V. 11

therefore the figure

*A*is similar to the figure

*B*. [VI.

*Def*. 1.

*rectilineal figures*&c. q.e.d.

PROPOSITION 22. *THEOREM*.

*If four straight lines he proportionals, the similar rec- tilineal figures similarly described on them shall also be proportionals; and if the similar rectilineal figures simi- larly described on four straight lines be proportionals, those straight lines shall be proportionals.*

Let the four straight lines *AB*, *CD*, *EF*, *GH* be proportionals, namely, *AB* to *CD as *EF* is to *GH*;*
and on

*AB*,

*CD*let the similar rectilineal figures

*KAB*,

*LCD*be similarly described;

and on

*EF*,

*GH*let the similar rectilineal figures

*MF*,

*NH*be similarly described:

the figure

*KAB*shall be to the figure

*LCD*as the figure

*MF*is to the figure

*NH*. To

*AB*and

*CD*take a third proportional

*X*, and to

*EF*and

*GH*a third proportional

*O*. [VI. 11.

Then, because

*AB*is to

*CD*as

*EF*is to

*GH*, [

*Hypothesis*.

and

*AB*is to

*CD*as

*CD*is to

*X*; [Construction.

and

*EF*is to

*GH*as

*GH*is to

*O*; [Construction.

therefore

*CD*is to

*X*as GH is to O. [V. 11.

And

*AB*is to

*CD*as

*EF*is to

*GH*;

therefore, ex aequali,

*AB*is to X as EF is to O. [V. 22.

But as

*AB*is to

*X*, so is the rectilineal figure

*KAB*to the rectilineal figure

*LCD*; [VI. 20,

*Corollary*2.

and as

*EF*is to

*O*, so is the rectilineal figure

*MF*to the rectilineal figure

*NH*; [VI. 20,

*Corollary*2.

therefore the figure

*KAB*is to the figure

*LCD*as the figure

*MF*is to the figure

*NH*. [V. 11.

Next, let the figure *KAB* be to the similar figure *LCD*
as the figure *MF* is to the similar figure *NH*; *AB* shall
be to *CD* as *EF* is to *GH*.

Make as *AB* is to *CD* so *EF* to *PR* : [VI. 12.

and on *PR* describe the rectilineal figure *SR*, similar and
similarly situated to either of the figures *MF*, *NH*. [VI, 18.

Then, because *AB* is to *CD* as *EF* is to *PR*,
and that on *AB*, *CD* are described the similar and simi-
larly situated rectilineal figures *KAB*, *LCD*,

and on *EF*, *PR* the similar and similarly situated recti-
lineal figures *MF*, *SR* ;

therefore, by the former part of this proposition, *KAB* is
to *LCD* as *MF* is to *SR*.

But, by hypothesis, *KAB* is to *LCD* as *MF* is to *NH*;

therefore *MF* is to *SR* as *MF* is to *NH* ; [V. 11.

therefore *SR* is equal to *NH*. [V. 9.

But the figures *SR* and *NH* are similar and similarly
situated, [*Construction*.

therefore *PR* is equal to *GH*.

And because *AB* is to *CD* as *EF* is to *PR*,

and that *PR* is equal to *GH* ;

therefore *AB* is to *CD* as *EF* is to *GH*. [V. 7.

Wherefore, *if four straight lines* &c. q.e.d.

*THEOREM*.

*Parallelograms with are equiangular to one another have to one another the ratio which is compounded of the ratios of their sides.* Let the parallelogram *AC* be equiangular to the parallelogram *CF*, having the angle *BCD* equal to the angle *ECG*: the parallelogram *AC* shall have to the parallelogram *CF* the ratio which is compounded of the ratios of their sides.
Let *BC* and *CG* be placed in a straight line;

therefore *DC* and *CE* are also in a straight line; [I. 1 4,

complete the parallelogram *DG*;

take any straight line *K*, and make *K* to *L* as *BC* is to *CG* and *L* to *M* as *DC* to *CE*; [VI 12.

then the ratios of *K* to *L* and of *L* to *M* are the same with the ratios of the sides, namely, of *BC* to *CG* and of *DC* to *CE*.

But the ratio of *K* to *M* is that which is said to be compounded of the ratios of *K* to *L* and of *L* to *M*; [V. *Def*. A.

therefore K has to M the ratio which is compounded of the ratios of the sides.

Now the parallelogram *AC* is to the parallelogram *CH* as *BC* is to *CG*; [VI. 1.

but BC is to CG as ^is to Z; [*Construction.*

therefore the parallelogram *AC* is to the parallelogram *CH* as *K*is to *L*. [V. 11.</br Again, the parallelogram *CH* is to the parallelogram *CF* as DC is to CE; [VI 1.

but *DC* is to *CE* as *L* is to *M*; [*Construction*.

therefore the parallelogram *CH* is to the parallelogram *CF* as *L* is to *M*. [V. 11.

Then, since it has been shewn that the parallelogram *AC* is to the parallelogram *CH* as *K *is to *L*,

and that the parallelogram *CH* is to the parallelogram *CF* as *L* is to *M*,

therefore, ex sequali, the parallelogram *AC* is to the parallelogram *CF* as *K* is to *M*. [V. 22. But *K* has to *M* the ratio which is compounded of the ratios of the sides;

therefore also the parallelogram *AC* has to the parallelogram *CF* the ratio which is compounded of the ratios of the sides.

Wherefore, *parallelograms* &c. q.e.d.

*THEOREM*.

*Parallelograms about the diameter of any parallelogram are similar to the whole parallelogram and to one another.*

Let *ABCD* be a parallelogram, of which *AC* is a diameter; and let *EG* and *HK* be parallelograms about the diameter: the parallelograms *EG* and *HK* shall be similar both to the whole parallelogram and to one another.

For, because DC and GF are parallels,

the angle *ADC* is equal to the angle *AGF*. [I. 29.

And because *BC* and *EF* are parallels, the angle *ABC* is equal to the angle *AEF*. [I. 29.

And each of the angles *BCD* and *EFG* is equal to the opposite angle *BAD*, [I. 34.

and therefore they are equal to one another.

Therefore the parallelograms *ABCD* and *AEFG* are equiangular to one another.

And because the angle *ABC* is equal to the angle *AEF*, and the angle *BAC* is common to the two triangles *BAC* and *EAF*,

therefore these triangles are equiangular to one another; and therefore *AB* is to *BC* as *AE* is to *EF*. [VI. 4.

And the opposite sides of parallelograms are equal to one another;

therefore *AB* is to *AD* as *AE* is to *AG*,

and *DC* is to *CB* as *GF* is to *FE*,

and *CD* is to *DA* as *FG* is to *GA*.[V.7

Therefore the sides of the parallelograms *ABCD* and *AEFG* about their equal angles are proportional,

and the parallelograms are therefore simillar to one another. [VI. *Definition* 1.

For the same reason the parallelogram *ABCD* is similar to the parallelogram *FHCK*.

Therefore each of the parallelograms *EG* and *HK* is similar to *BD*;

therefore the parallelogram *EG* is similar to the parallelogram *HK'.*

Wherefore, *parallelograms* &c. q.e.d.

*PROBLEM*.

*To describe a rectilineal figure which shall he similar to one given rectilineal figure and equal to another given rectilineal figure.*

Let *ABC* be the given rectilineal figure to which the figure to be described is to be similar, and *D* that to which it is to be equal: it is required to describe a rectilineal figure similar to *ABC* and equal to *D*.

On the straight line *BC* describe the parallelogram *BE* equal to the figure *ABC*.

On the straight line *CE' describe the parallelogram *CM* equal to *D*, and having the angle *FCE* equal to the angle CBL; [I. 45, *Corollary*. therefore *BC' and

*CF*will be in one straight line, and

*LE*and

*EM*will be in one straight line.

Between

*BC*and

*CF*find a mean proportional

*GH*, [VI. 13.

and on GH describe the rectilineal figure KGH, similar and similarly situated to the rectilineal figure ABC. [VI. 18.

*KGH*shall be the rectilineal figure required.

For, because *BG* is to *GH* as *GH* is to *GF*, [Construction.

and that if three straight lines be proportionals, as the first is to the third so is any figure on the first to a similar and similarly described figure on the second, [VI. 20, *Cor*. 2.

therefore as *BC* is to *GF* so is the figure *ABC* to the figure KGH.

But as *BC* is to *CF* so is the parallelogram *BE* to the parallelogram *CM*; [VI. 1.

therefore the figure ABG is to the figure KGH as the parallelogram BE is to the parallelogram GM. [V. 11.

And the figure ABG is equal to the parallelogram BE;

therefore the rectilineal figure KGH is equal to the parallelogram GM. [V. 14.

But the parallelogram GM is equal to the figure D; [*Constr.*

therefore the figure KGH is equal to the figure D, [Axiom 1.

and it is similar to the figure ABG. [Construction.

Wherefore the *rectilineal figure KGH has heen de-scribed similar to the figure ABC, and equal to D*. q.e.f.

*THEOREM*.

*If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.*

Let the parallelograms *ABCD*, *AEFG* be similar and similarly situated, and have the common angle *BAD*: *ABCD* and *AEFG* shall be about the same diameter.

For, if not, let, if possible, the parallelogram *BD* have its diameter *AHC* in a different straight line from *AF*, the diameter of the prallelogram *EG*; let *GF* meet *AHC* at *H*, and through *H* draw *HK* parallel to *AD or BC. [1. 31.*

Then the parallelograms *ABCD* and *AKHG* are about the same diameter, and are therefore similar to one another; [VI. 24.

therefore *DA* is to *AB* as *GA* is to *AK*.

But because *ABCD* and *AEFG* are similar parallelograms, [*Hypothesis*.

therefore *DA* is to *AB* as *GA* is to *AE*. [VI. Definition 1.

Therefore *GA* is to *AK* as *GA* is to *AE*, [V. 11. that is, *GA* has the same ratio to each of the straight lines *AK* and *AE*,

and therefore *AK* is equal to *AE*, [V. 9.

the less to the greater; which is impossible.

Therefore the parallelograms *ABCD* and *AEFG* must have their diameters in the same straight line, that is, they are about the same diameter.

Wherefore, if *two similar parallelograms* &c. q.e.d.

*PROBLEM*.

*To cut a given straight line in extreme and mean ratio.*

Let *AB* be the given straight line: it is required to cut it in extreme and mean ratio.

Divide *AB* at the point *C*, so that the rectangle contained by *AB*, *BC* may be equal to the square on AC. [II.11.

Then, because the rectangle *AB*, *BC* is equal to the square on *AC*, [*Construction*.

therefore AB is to AC a,s AC is to CB. [VI. 17.

*Wherefore*AB

*is cut in extreme and mean ratio at the point*C. q.e.f. [VI.

*Definition*3.

*THEOREM*.

*In any right-angled triangle, any rectilineal figure described on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle*.

Let *ABC* be a right-angled triangle, having the right angle *BAC*: the rectilineal figure described on *BC* shall be equal to the similar and similarly described figures on *BA* and *CA*.
Draw the perpendicular *AD*. [I. 12.

Then, because in the rightangled triangle *ABC*, *AD* is drawn from the right angle at *A*, perpendicular to the base *BC*, the triangles *ABD*, *CAD* are similar to the whole triangle *CBA*,and to one another. [VI. 8.

And because the triangle *CBA* is similar to the triangle *ABD*,

therefore *CB* is to *BA* as *BA* is to *BD*. [VI. *Def*. 1.

And when three straight lines are proportionals, as the first is to the third so is the figure described on the first to the similar and similarly described figure on the second; [VI. 20, *Corollary* 2.

therefore as *CB* is to *BD* so is the figure described on *CB* to the similar and similarly described figure on *BA*;

and inversely, as *BD* is to *BC* so is the figure described on *BA* to that described on *CB*. [V. B.

In the same manner, as *CD* is to *CB* so is the figure described on *CA* to the similar figure described on *CB*.

Therefore as *BD* and *CD* together are to *CB* so are the figures described on *BA* and *CA* together to the figure described on *CB*. [V. 24.

Cut BD and CD together are equal to CB;

therefore the figure described on BC is equal to the similar and similarly described figures on *BA* and *CA*. [V. *A*.

*in any right-angled triangle*&c. q.e.d.

*THEOREM*.

*If two triangles, which have two sides of the one pro- portional to two sides of the other, he joined at one angle so as to have their homologous sides parallel to one another, the remaining sides shall he in a straight line*.

Let *ABC* and *DCE* be two triangles, which have the two sides *BA*, *AC* proportional to the two sides *CD*, *DE*, namely, *BA* to *AC* as *CD* is to *DE* and let *AB* be parallel to *DC* and *AC* parallel to *DE*: *BC* and *CE* shall be in one straight line.

For, because *AB* is parallel to *DC*, [*Hypothesis*.

and *AC* meets them, the alternate angles *BAC*, *ACD* are equal; [I. 29.

for the same reason the angles *ACD*, *CDE* are equal;

therefore the angle *BAC* is equal to the angle *CDE*. [*Ax*. 1.

And because the triangles *ABC*, *DCE* have the angle at *A* equal to the angle at *D*, and the sides about these angles proportionals, namely, *BA* to *AC* as *CD* is to *DE*, [*Hyp*.

therefore the triangle *ABC* is equiangular to the triangle *DCE*; [VI. 6.

therefore the angle *ABC* is equal to the angle *DCE*.

And the angle *BAC* was shewn equal to the angle *ACD*;

therefore the whole angle *ACE* is equal to the two angles *ABC* and *BAC*. [*Axiom* 2.

Add the angle *ACB* to each of these equals;

then the angles *ACE* and *ACB* are together equal to the angles *ABC*, *BAC*, *ACB*.

But the angles *ABC, *BAC*, *ACB* are together equal to two right angles; [I. 32.*
therefore the angles

*ACE*and

*ACB*are together equal to two right angles.

And since at the point *C*, in the straight line *AC*, the two straight lines *BC*, *CE* which are on the opposite sides of it, make the adjacent angles *ACE*,*ACB* together equal to two right angles,

therefore *BC* and *CE* are in one straight line. [I. 14.

Wherefore, *if two triangles* &c. q.e.d.

*THEOREM*.

*In equal circles, angles, whether at the centres or at the circumferences, have the same ratio which the arcs on which they stand have to one another; so also have the sectors.*

Let '*ABC* and *BEF* be equal circles, and let *BGC* and *EHF* be angles at their centres, and *BAC* and *EDF* angles at their circumferences: as the arc *BC* is to the arc *EF* so shall the angle *BGC* be to the angle *EHF*, and the angle *BAC* to the angle *EDF*; and so also shall the sector *BGC* be to the sector *EHF*.

Take any number of arcs *CK*, *KL*, each equal to *BC*, and also any number of arcs *FM*, *MN* each equal to *EF*; and join *GK*, *GL*, *HM*, *HN*.

Then, because the arcs *BC*, *CK*, *KL*, are all equal, [*Constr*.

the angles *BCG*, *CGK*, *KGL* are also all equal; [III. 27.

and therefore whatever multiple the arc *BL* is of the arc *BC*, the same multiple is the angle *BGL* of the angle *BGC*.

For the same reason, whatever multiple the arc *EN* is of the arc *EF*, the same multiple is the angle *EHN* of the angle *EHF*.

And if the arc *BL* be equal to the arc EN, the angle BGL is equal to the angle EHN; [III. 27.

and if the arc *BL* be greater than the arc *EN*, the angle *BGL* is greater than the angle *EHN*; and if less, less.

Therefore since there are four magnitudes, the two arcs *BC*,*EF*, and the two angles *BGC*, *EHF*;
and that of the arc *BC* and of the angle *BGC* have been taken any equimultiples whatever, namely, the arc *BL* and the angle *BGL*;

and of the arc *EF* and of the angle *EHF* have been taken any equimultiples whatever, namely, the arc *EN* and the *EHN*;

and since it has been shewn that if the arc *BL* be greater than the arc *EN*, the angle *BGL* is greater than the angle *EHN*; and if equal, equal; and if less, less;

therefore as the arc *BC* is to the arc *EF*, so is the angle *BGC* to the angle *EHF*. [V. Definition 5.

But as the angle *BGC* is to the angle *EHF*, so is the angle *BAC* to the angle *EDF*, [V. 15.

for each is double of each; [III. 20.

therefore, as the arc *BC* is to the arc *EF* so is the angle *BGC* to the angle *EHF*, and the angle *BAC* to the angle *EBF*.

Also as the arc *BC* is to the arc *EF*, so shall the sector *BGC* be to the sector *EHF*.

Join *BC*, *CK*, and in the arcs *BC*, *CK* take any points *X*, *O*, and join *BX*, *XC*, *CO*, *OK*.

Then, because in the triangles *BGC*,*CGK* the two sides *BG*, *GC* are equal to the two sides *CG*, *GK*, each to each;

and that they contain equal angles; [III. 27.

therefore the base *BC* is equal to the base *CK*, and the triangle *BGC* is equal to the triangle *CGK*, [I. 4.

And because the arc *BC* is equal to the arc *CK*, [*Constr*.

the remaining part when *BC* is taken from the circumference is equal to the remaining part when *CK* is taken from the circumference;

therefore the angle *BXC* is equal to the angle *COK*. [III. 27.

Therefore the segment *BXC* is similar to the segment *COK*; [III. *Definition* 11.

and they are on equal straight lines *BC*, *CK*.

But similar segments of circles on equal straight lines are equal to one another; [III. 24.

therefore the segment *BXC* is equal to the segment *COK*.

And the triangle *BGC* was shewn to be equal to the triangle *CGK*;

therefore the whole, the sector *BGC*, is equal to the whole, the sector *CGK* [*Axiom* 2.

For the same reason the sector *KGL* is equal to each of the sectors *BGC*, *CGK*,

In the same manner the sectors *EHF*, *FHM*, *MHN* may be shewn to be equal to one another.

Therefore whatever multiple the arc *BL* is of the arc *BC*, the same multiple is the sector *BGL* of the sector *BGC*;

and for the same reason whatever multiple the arc *EN* is of the arc *EF* the same multiple is the sector *EHN* of the sector *EHF*.

And if the arc *BL* be equal to the arc *EN* the sector *BGL* is equal to the sector *EHN*;

and if the arc *BL* be greater than the arc *EN* the sector *BGL* is greater than the sector *EHN*; and if less, less.

Therefore, since there are four magnitudes, the two arcs *BC*, *EF*, and the two sectors *BGC*, *EHF*;

and that of the arc *BC* and of the sector *BGC* have been taken any equimultiples whatever, namely, the arc *BL* and the sector *BGL*;

and of the arc *EF* and of the sector *EHF* have been taken any equimultiples whatever, namely, the arc *EN* and the sector *BGL*;

and since it has been shewn that if the arc *BL* be greater than the arc *EN*, the sector *BGL* is greater than the sector *EHN*; and if equal, equal; and if less, less;

therefore as the arc *BC* is to the arc *EF*, so is the sector *BGC* to the sector *EHF*. [V. *Definition* 5.

*in equal circles*&c. q.e.d.

*B*.

*THEOREM*.

*If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle con-tained by the segments of the base, together with the square on the straight line which bisects the angle*.

Let *ABC* be a triangle, and let the angle *BAC* be bisected by the straight line *AD*: the rectangle *BA*, *AC* shall be equal to the rectangle *BD*,*DC*, together with the square on *AD*.

Describe the circle *ACB* about the triangle, [IV. 5.

and produce AD to meet the circumference at *E*,

and join *EC*.

Then, because the angle *BAD* is equal to the angle *EAC*, [*Hypothesis*.

and the angle *ABD* is equal to the angle *ABC*, for they are in the same segment of the circle, [III. 21.

therefore the triangle *BAD* is equiangular to the triangle *EAC*.

Therefore *BA* is to *AD* as *BA* is to *AC*; [VI. 4.

therefore the rectangle *BA*, *AC* is equal to the rectangle *EA*,*AD*, [VI. 16.

that is, to the rectangle *BD*, *DA*, together with the square on *AD*. [II. 3,

But the rectangle JSD, DA is equal to the rectangle *BD*,*DC*; [III.35.

therefore the rectangle *BA*, *AC* is equal to the rectangle *BD*, *DC*, together with the square on *AD*.

*if the vertical angle*&c. q.e.d.

*C*.

*THEOREM*.

*If from the vertical angle of a triangle a straight line, be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle con-tained by the perpendicular and the diameter of the circle described about the triangle.*

Let *ABC* be a triangle, and let *AD* be the perpendicular from the angle *A* to the base *BC* the rectangle *BA*, *AC* shall be equal to the rectangle contained by *AD* and the diameter of the circle described about the triangle.

Describe the circle *ACB* about the triangle; [IV. 5.

draw the diameter *AE*, and join *EC*

Then, because the right angle *BDA* is equal to the angle *ECA* in a semi-circle; [III. 31.

and the angle *ABD* is equal to the angle *AEC*, for they are in the same segment of the circle; [III. 21.

therefore the triangle *ABD* is equiangular to the triangle *AEC'.*
Therefore

*BA*is to

*AD*as

*EA*is to

*AC*; [VI; 4.

therefore the rectangle

*BA*,

*AC*is equal to the rectangle

*EA*,

*AD*. [ VI. 16.

Wherefore, *if from the vertical angle* &c. q.e.d.

*D*.

*THEOREM*.

*The rectangle contained by the diagonals of a quadri-lateral figure inscribed in a circle is equal to both the rectangles contained by its opposite sides.* Let *ABCD* be any quadrilateral figure inscribed in a circle, and join *AC*,*BD*: the rectangle contained by *AC*, *BD* shall be equal to the two rectangles contained by *AB*,*CD* and *AD*,*BC*.

Make the angle *ABE* equal to the angle *DBC*; [I. 23.

add to each of these equals the angle *EBD*,

then the angle *ABD* is equal to the angle *EBC. [*Axiom* 2.*
And the angle

*BDA*is equal to the angle

*BCE*, for they are in the same segment of the circle; [III.21.

therefore the triangle

*ABD*is equiangular to the triangle

*EBC*.

Therefore

*AD*is to

*DB*as

*EC*is to

*CB*; [VI. 4.

therefore the rectangle

*AD*,

*CB*is equal to the rectangle

*DB*, EC. [VI. 16.

Again, because the angle *ABE* is equal to the angle *DBC*, [*Construction*.

and the angle *BAE* is equal to the angle *BDC, for they are in the same segment of the circle; [III. 21.*
therefore the triangle

*ABE*is equiangular to the triangle

*DBC*.

Therefore

*BA*is to

*AE*as

*BD*is to DC; [VI. 4.

therefore the rectangle

*BA*,

*DC*is equal to the rectangle

*AE*,

*BD*. [VI. 16.

But the rectangle *AD*, *CB* has been shewn equal to the rectangle *DB*, *EC*;

therefore the rectangles *AD*, *CB* and *BA*, *DC* are together equal to the rectangles *BD*, *EC* and *BD*, *AE*;

that is, to the rectangle *BD*, *AC*. [II. 1.

Wherefore, *the rectangle contained* &c. q.e.d.