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

 * Euclid-II-11.png

Let $AB$ be the given straight line segment.

Construct the square $ABDC$ on $AB$.

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

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

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.