# Construction of Square equal to Given Polygon

## Theorem

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

In the words of Euclid:

*To construct a square equal to a given rectilineal figure.*

(*The Elements*: Book $\text{II}$: Proposition $14$)

## Proof

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.

$\blacksquare$

## Historical Note

This proof is Proposition $14$ of Book $\text{II}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{II}$. Propositions