Triangles with Two Sides Parallel and Equal

Proof
Let $\triangle ABC, \triangle DCE$ be two triangles such that $BA : AC = DC : DE$ be situated so that $AB \parallel DC$ and $AC \parallel DE$.

We need to show that $BC$ is in a straight line with $CE$.


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We have that $AB \parallel DC$.

So from Parallelism implies Equal Alternate Angles we have that $\angle BAC = \angle ACD$.

For the same reason $\angle CDE = \angle ACD$.

So $\angle BAC = \angle CDE$.

From Triangles with One Equal Angle and Two Sides Proportional are Similar, it follows that $\triangle ABC$ is similar to $\triangle DCE$.

Therefore $\angle ABC = \angle DCE$.

It follows that $\angle ACE = \angle ABC + \angle BAC$.

Add $\angle ACB$ to each.

Then $\angle ACE + \angle ACB = \angle BAC + \angle ACB + \angle CBA$.

But from Sum of Angles of Triangle Equals Two Right Angles $\angle BAC + \angle ACB + \angle CBA$ equals two right angles.

So $\angle ACE + \angle ACB$ equals two right angles.

The result follows from Two Angles making Two Right Angles make Straight Line.