Construction of Parallelogram equal to Triangle in Given Angle
Theorem
A parallelogram can be constructed in a given angle the same size as any given triangle.
In the words of Euclid:
- To construct, in a given angle, a parallelogram equal to a given triangle.
(The Elements: Book $\text{I}$: Proposition $42$)
Proof
Let $ABC$ be the given triangle, and $D$ the given angle.
Bisect $BC$ at $E$, and join $AE$.
Construct $AG$ parallel to $EC$.
Construct $\angle CEF$ equal to $\angle D$.
Construct $CG$ parallel to $EF$.
Then $FEGC$ is a parallelogram.
Since $BE = EC$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle ABE = \triangle AEC$.
So $\triangle ABC$ is twice the area of $\triangle AEC$.
But from Parallelogram on Same Base as Triangle has Twice its Area, $FECG$ is also twice the area of $\triangle AEC$.
So $FECG$ has the same area as $\triangle ABC$, and has the given angle $D$.
$\blacksquare$
Historical Note
This proof is Proposition $42$ of Book $\text{I}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions