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

 * Euclid-I-42.png

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 $$D$$.