Construction of Parallelogram Equal to Given Figure Less a Parallelogram

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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 $D$.

Euclid-VI-28.png

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

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