Construction of Square equal to Given Polygon

Theorem
A square can be constructed the same size as any given polygon.

Proof

 * Euclid-II-14.png

Let $$A$$ be the given polygon.

Construct the rectangle $BCDE$ equal to the given polygon.

If it so happens that $$BE = ED$$, then $$BCDE$$ is a square, and the construction is complete.

Suppose $$BE \ne ED$$. Then WLOG suppose $$BE > ED$$.

Produce $BE$ from $$E$$ and construct on it $EF = ED$.

Bisect $BF$ at $$G$$.

Construct the semicircle $$BHF$$ with center $G$ and radius $GF$ (see diagram).

Produce $DE$ from $$D$$ to $$H$$.

From Difference of Two Squares, the rectangle contained by $$BE$$ and $$EF$$ together with the square on $$EG$$ is equal to the square on $$GF$$.

But $$GF = GH$$.

So the rectangle contained by $$BE$$ and $$EF$$ together with the square on $$EG$$ is equal to the square on $$GH$$.

From Pythagoras's Theorem, the square on $$GH$$ equals the squares on $$GE$$ and $$EH$$.

Then the rectangle contained by $$BE$$ and $$EF$$ together with the square on $$EG$$ is equal to the squares on $$GE$$ and $$EH$$.

Subtract the square on $$GE$$ from each.

Then the rectangle contained by $$BE$$ and $$EF$$ is equal to the square on $$EH$$.

So the square on $$EH$$ is equal to the rectangle $$BCDE$$.

So the square on $$EH$$ is equal to the given polygon, as required.