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.
PROPOSITION 5. 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.
