*EUCLID'S ELEMENTS.*

*bases equal, the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides, equal to them, of the other.*

Let *ABC*, *DEF* be two triangles, having the two sides *AB*, *AC* equal to the two sides *DE*, *DF*, each to each, namely *AB* to *DE*, and *AC* to *DF*, and also the base *BC* equal to the base *EF*: the angle *BAC* shall be equal to the angle *EDF*.

For if the triangle *ABC* be applied to the triangle *DEF*, so that the point *B* may be on the point *E*, and the straight line *BC* on the straight line *EF*, the point *C* will also coincide with the point *F*, because *BC* is equal to *EF*. [*Hyp*.

Therefore, *BC* coinciding with *EF*, *BA* and *AC* will coincide with *ED* and *DF*.

For if the base *BC* coincides with the base *EF*, but the sides *BA*, *CA* do not coincide with the sides *ED*, *FD*, but have a different situation as *EG*, *FG*; then on the same base and on the same side of it there will be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise their sides which are terminated at the other extremity.

But this is impossible. [I. 7.

Therefore since the base *BC* coincides with the base *EF*, the sides *BA*, *AC* must coincide with the sides *ED*, *DF*. Therefore also the angle *BAC* coincides with the angle *EDF*, and is equal to it. [*Axiom* 8.

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

PROPOSITION 9. *PROBLEM*.

To bisect a given rectilineal angle, that is to divide it into two equal angles.