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.