Construction of Square equal to Rectangle
Theorem
In the words of Euclid:
- To cut a given a straight line so that the rectangle contained by the whole and one of the segments equals the square on the remaining segments.
(The Elements: Book $\text{II}$: Proposition $11$)
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 proof is Proposition $11$ 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