# Construction of Parallelogram on Given Line equal to Triangle in Given Angle

## Contents

## Theorem

In the words of Euclid:

*To a given straight line to apply, in a given angle, a parallelogram equal to a given triangle.*

(*The Elements*: Book $\text{I}$: Proposition $44$)

## Proof

Let $AB$ be the given line segment, $C$ be the given triangle and let $D$ be the given angle.

From Construction of Parallelogram equal to Triangle in Given Angle, construct the parallelogram $BEFG$ equal to $C$ such that $\angle EBG = \angle D$.

Place it so that $BE$ is in a straight line with $AB$.

Let $FG$ be produced to $H$, and draw $AH$ parallel to $BG$.

Then join $HB$.

We have that $HF$ falls on the parallel lines $AH$ and $EF$.

So from Parallelism implies Supplementary Interior Angles, $\angle AHF$ and $\angle HFE$ are supplementary.

So $\angle BHG + \angle GFE$ is less than two right angles.

From the Fifth Postulate, $HB$ and $FE$ will meet if produced. So let them meet at $K$.

Draw $KL$ parallel to $EA$, and produce $HA$ and $GB$ to the points $L$ and $M$.

Then $HLKF$ is a parallelogram.

From Complements of Parallelograms are Equal, $ABML$ has the same area as $BEFG$.

But $BEFG$ has the same area as $\triangle C$.

So by common notion 1, so does $ABML$.

From Two Straight Lines make Equal Opposite Angles, $\angle GBE = \angle ABM$, while $\angle GBE = \angle D$.

Therefore $ABML$ is the required parallelogram.

$\blacksquare$

## Historical Note

This theorem is Proposition $44$ of Book $\text{I}$ of Euclid's *The Elements*.

There is a suggestion that this result is the work of Pythagoras.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions