Construction of Parallelogram Equal to Given Figure Less a Parallelogram
Theorem
In the words of Euclid:
- To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.
(The Elements: Book $\text{VI}$: Proposition $28$)
Construction
Let:
- $AB$ be the given straight line
- $C$ be the given rectilineal figure to which the figure to be applied to $AB$ is to be equal, not being greater than the parallelogram described on the half of $AB$ and similar to the defect
and:
- $D$ be the parallelogram to which the defect required is to be similar.
We are to construct a parallelogram on $AB$ equal to $C$ and deficient by a parallelogrammic figure similar to $D$.
Let $AB$ be bisected at $E$.
Using Construction of Similar Polygon, let $\Box EBFG$ be described similar to $D$ and similarly situated.
Let the parallelogram $\Box AG$ be completed.
If $AG = C$ then the job is done.
Otherwise let $\Box HE > C$.
By Construction of Figure Similar to One and Equal to Another, let $\Box KLMN$ be constructed equal to the excess by which $GB$ is greater than $C$ and similar and similarly situated to $D$.
Let $GO = KL$ and $GP = LM$, and complete the parallelogram $\Box OGPQ$.
Construct the lines $RT \parallel AB$ and $PS \parallel GE$ passing through $Q$.
Then the parallelogram $ST$ is the required figure.
Proof
Let $AG = C$.
Then $\Box AG$ has been applied to $AB$ equal in area to $C$ and deficient by $\Box GB$ similar to $D$.
Otherwise let $\Box HE > C$.
Now $HE = GB$, so:
- $GB > C$
We have that $\Box KLMN$ is constructed equal to the excess by which $GB$ is greater than $C$ and similar and similarly situated to $D$.
But $D$ is similar to $GB$ and so from Similarity of Polygons is Equivalence Relation $KM$ is also similar to $GB$.
Let $KL$ correspond to $GE$ and $LM$ to $GF$.
Since $\Box GB = C + \Box KLMN$ it follows that:
- $\Box GB > \Box KLMN$
Therefore $GE > KL$ and $GF > KM$.
We have $GO$ constructed equal to $KL$ and $GP$ equal to $KM$.
Therefore $\Box OGPQ = \Box KLMN$ and $\Box OGPQ$ is similar to $\Box KLMN$.
Therefore from Similarity of Polygons is Equivalence Relation $\Box GQ$ is similar to $\Box GB$.
So from Parallelogram Similar and in Same Angle has Same Diameter, $GQ$ is about the same diameter with $GB$.
Let $GQB$ be this diameter and let the figure be described.
Since $\Box BG = C + \Box KM$, and in them $\Box GQ = \Box KM$, it follows that the gnomon $UWV$ equals the remainder $C$.
From [[Complements of Parallelograms are Equal]:
- $\Box PR = \Box OS$.
Let $\Box QB$ be added to each.
Then:
- $\Box PB = \Box OB$
But since $AE = EB$, it follows from Parallelograms with Equal Base and Same Height have Equal Area that:
- $\Box OB = \Box TE$
Therefore:
- $\Box TE = \Box PB$
Let $\Box OS$ be added to each.
Therefore the whole of $\Box TS$ equals the area of the gnomon $UWV$.
But $UWV$ was shown to be equal to $C$.
Hence the result.
$\blacksquare$
Historical Note
This proof is Proposition $28$ of Book $\text{VI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions