# Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram

## Theorem

In the words of Euclid:

*To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.*

(*The Elements*: Book $\text{VI}$: Proposition $29$)

## Construction

Let $AB$ be the given straight line, $C$ the given rectilineal figure to which the figure to be applied to $AB$ is to be equal, and $\Box D$ that to which the excess is required to be similar.

We need to apply to $AB$ a parallelogram equal to the area of $C$ exceeding it by a parallelogrammic figure similar to $\Box D$.

Let $AB$ be bisected at $E$.

Describe on $EB$ the parallelogram $\Box BF$ similar and similarly situated to $\Box D$.

From Construction of Figure Similar to One and Equal to Another, let $\Box GH$ be constructed at once equal to $\Box BF + C$ and similarly situated to $D$.

Let $FL$ and $FE$ be produced to $M$ and $N$ so that $FLM = KH$ and $FEN = KG$, and complete the parallelogram $\Box MN$, whose diameter is $FO$.

Construct the parallelogram $\Box AO$.

Then $\Box AO$ is the required parallelogram.

## Proof

Let $KH$ correspond to $FL$ and $KG$ to $FE$.

Since $GH > FB$, we have that $KH > FL$ and $KG > FE$.

Since $FLM = KH$ and $FEN = KG$, it follows that $\Box MN = \Box GH$ and also $\Box MN$ is similar to $\Box GH$.

But $\Box GH$ is similar to $\Box EL$.

So from Similarity of Polygons is Equivalence Relation, $\Box MN$ is similar to $\Box EL$.

Therefore from Parallelogram Similar and in Same Angle has Same Diameter $\Box EL$ is about the same diameter with $\Box MN$.

Let that diameter $FO$ be drawn, and the figure described.

Since $\Box GH = \Box EL + C$, while $\Box MN = \Box GH$, it follows that $\Box MN = \Box EL + C$.

Subtract $\Box EL$ from each.

Then the gnomon $XWV$ is equal to $C$.

Since $AE = EB$, from Parallelograms with Equal Base and Same Height have Equal Area, $\Box AN = \Box NB$.

Thus, from Complements of Parallelograms are Equal, $\Box AN = \Box LP$.

Add $\Box EO$ to each.

Then $\Box AO$ equals the gnomon $XWV$.

But the gnomon $XWV$ equals $C$.

So $\Box AO = C$.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $29$ 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