# Construction of Parallelogram in Given Angle equal to Given Polygon

## Theorem

A parallelogram can be constructed in a given angle the same size as any given polygon.

In the words of Euclid:

*To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.*

(*The Elements*: Book $\text{I}$: Proposition $45$)

## Proof

Let $ABCD$ be the given polygon, and let $E$ be the given angle.

Join $DB$, and construct the parallelogram $FGHK$ equal in size to $\triangle ABD$, in $\angle HKF = \angle E$.

Then construct the parallelogram $GLMH$ equal in area to $\triangle BCD$ on the line segment $GH$, in $\angle GHM = \angle E$.

We now need to show that $KFLM$ is the required parallelogram.

By Common Notion $1$, $\angle HKF = \angle GHM$ as both are equal to $\angle E$.

Add $\angle KHG$ to each, so as to make $\angle FKH + \angle KHG = \angle KHG + \angle GHM$.

From Parallelism implies Supplementary Interior Angles:

- $\angle FKH + \angle KHG$

Therefore $\angle KHG + \angle GHM$ equal two right angles.

So from Two Angles making Two Right Angles make Straight Line, $KH$ is in a straight line with $HM$.

From Parallelism implies Equal Alternate Angles:

- $\angle MHG = \angle HGF$

Add $\angle HGL$ to each, so as to make $\angle MHG + \angle HGL = \angle HGF + \angle HGL$.

From Parallelism implies Supplementary Interior Angles:

- $\angle MHG + \angle HGL$

Therefore $\angle HGF + \angle HGL$ equal two right angles.

So from Two Angles making Two Right Angles make Straight Line, $FG$ is in a straight line with $GL$.

From Parallelism is Transitive Relation, as $KF \parallel HG$ and $HG \parallel ML$, it follows that $KF \parallel ML$.

Similarly, from common notion 1, $KF = ML$.

As $KM$ and $FL$ join them at their endpoints, $KM \parallel FL$ and $KM = FL$ from Lines Joining Equal and Parallel Straight Lines are Parallel.

Therefore $KFLM$ is a parallelogram.

But the area of $KFLM$ equals the combined areas of $FGHK$ and $GLMH$, which are equal to the combined areas of $\triangle ABD$ and $\triangle BCD$.

Therefore from Common Notion $2$, $KFLM$ has the same area as the polygon $ABCD$, in the angle $E$

$\blacksquare$

## Historical Note

This proof is Proposition $45$ of Book $\text{I}$ of Euclid's *The Elements*.

Note that this technique can be expanded for a polygon with any number of sides, merely by dividing the polygon up into as many triangles as it takes.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions - 1944: R.P. Gillespie:
*Integration*(2nd ed.) ... (previous) ... (next): Chapter $\text I$: $\S 1$. Area of a Circle