Construction of Square Equal to Rectangle

Theorem

Given a straight line segment, it is possible to cut it so that the rectangle contained by the whole and one of the segments equals the square on the remaining segment.

Proof

Let $AB$ be the given straight line segment.

Construct the square $ABDC$ on $AB$.

Bisect $CA$ at $E$ and join $BE$.

Produce $CA$ to $F$ and make $EF = BE$.

Construct the square $AFGH$ on $AF$.

Produce $GH$ to $K$.

Then $H$ is the point at which $AB$ has been cut so as to make the rectangle contained by $AB$ and $BH$ is the same size as the square on $AH$.

The proof is as follows.

From Square of Sum less Square, the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $EF$.

But $EF = EB$, so the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the square on $BE$.

By Pythagoras's Theorem, the squares on $AB$ and $AE$ equal the square on $BE$, because $\angle EAB$ is a right angle.

So the rectangle contained by $CF$ and $FA$ together with the square on $AE$ equals the squares on $AB$ and $AE$.

Subtract the square $AE$ from each.

Then the rectangle contained by $CF$ and $FA$ equals the square on $AB$.

Now the rectangle contained by $CF$ and $FA$ is $\Box CFGK$ because $AF = FG$.

Also, the square on $AB$ is $\Box ABDC$.

So $\Box CFGK = \Box ABDC$.

Subtract $AHKC$ from each.

Then $\Box FGHA = \Box HBDK$.

Now $\Box HBDK$ is the rectangle contained by $AB$ and $BH$, because $AB = BD$.

Also, $\Box FGHA$ is the square on $AH$.

Hence the result.

$\blacksquare$

Historical Note

This is Proposition 11 of Book II of Euclid's The Elements.