Construction of Parallelogram Equal to Triangle in Given Angle

Theorem

A parallelogram can be constructed in a given angle the same size as any given triangle.

Proof

Let $ABC$ be the given triangle, and $D$ the given angle.

Bisect $BC$ at $E$, and join $AE$.

Construct $AG$ parallel to $EC$.

Construct $\angle CEF$ equal to $\angle D$.

Construct $CG$ parallel to $EF$.

Then $FEGC$ is a parallelogram.

Since $BE = EC$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle ABE = \triangle AEC$.

So $\triangle ABC$ is twice the area of $\triangle AEC$.

But from Parallelogram on Same Base as Triangle has Twice its Area, $FECG$ is also twice the area of $\triangle AEC$.

So $FECG$ has the same area as $\triangle ABC$, and has the given angle $D$.

$\blacksquare$

Historical Note

This is Proposition 42 of Book I of Euclid's The Elements.