Triangles with Two Sides Parallel and Equal

Theorem

In the words of Euclid:

If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line.

(The Elements: Book $\text{VI}$: Proposition $32$)

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 $CE$.

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.

$\blacksquare$

Historical Note

This proof is Proposition $32$ of Book $\text{VI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions