Construction of Parallelogram in Given Angle equal to Given Polygon

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

Proof

 * Euclid-I-45.png

Let $$ABCD$$ be the given polygon, and let $$E$$ be the given angle.

Join $$DB$$, and construct the parallelogram $$FGHK$$ equal in size to $$\triangle ABD$$, in $$\angle HKF = \angle E$$.

Then construct the parallelogram $$GLMH$$ equal in area to $$\triangle BCD$$ on the line segment $$GH$$, in $$\angle GHM = \angle E$$.

We now need to show that $$KFLM$$ is the required parallelogram.

By common notion 1, $$\angle HKF = \angle GHM$$ as both are equal to $$\angle E$$.

Add $$\angle KHG$$ to each, so as to make $$\angle FKH + \angle KHG = \angle KHG + \angle GHM$$.

From Parallel Implies Supplementary Interior Angles, $$\angle FKH + \angle KHG$$ and therefore $$\angle KHG + \angle GHM$$ equal two right angles.

So from Two Angles making Two Right Angles make a Straight Line, $$KH$$ is in a straight line with $$HM$$.

From Parallel Implies Equal Alternate Interior Angles, $$\angle MHG = \angle HGF$$.

Add $$\angle HGL$$ to each, so as to make $$\angle MHG + \angle HGL = \angle HGF + \angle HGL$$.

From Parallel Implies Supplementary Interior Angles, $$\angle MHG + \angle HGL$$ and therefore $$\angle HGF + \angle HGL$$ equal two right angles.

So from Two Angles making Two Right Angles make a Straight Line, $$FG$$ is in a straight line with $$GL$$.

From Parallelism is Transitive, as $$KF \| HG$$ and $$HG \| ML$$, it follows that $$KF \| ML$$.

Similarly, from common notion 1, $$KF = ML$$.

As $$KM$$ and $$FL$$ join them at their endpoints, $$KM \| FL$$ and $$KM = FL$$ from Lines Joining Equal and Parallel Straight Lines.

Therefore $$KFLM$$ is a parallelogram.

But the area of $$KFLM$$ equals the combined areas of $$FGHK$$ and $$GLMH$$, which are equal to the combined areas of $$\triangle ABD$$ and $$\triangle BCD$$.

Therefore from common notion 2, $$KFLM$$ has the same area as the polygon $$ABCD$$, in the angle $$E$$