Construction of Equal Angle

Construction

 * Angle Construction.png

Let $A$ be the given point on the given straight line $AB$, and let $\angle DCE$ be the given rectilinear angle, in which points $D$ and $E$ are any points on the straight lines bounding the angle (one on each side).

We can then construct $\triangle AFG$ such that $CE = AF$, $CD = AG$, and $DE = GF$, with $F$ on $AB$.

$\angle GAF$ is the required angle.

Proof
Since all three sides of the triangles are equal, the interior angles of the triangles are also equal.

Thus, $\angle GAF = \angle ECD$, with $\angle GAF$ at the point $A$ on the straight line $AB$.

This theorem is one of the two attributed to (the other being ).