PROPOSITION 32. 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
