Construction of Equal Angle

Theorem

In the words of Euclid:

On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle.

(The Elements: Book $\text{I}$: Proposition $23$)

Construction

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 $AB$.

$\blacksquare$

Historical Note

This proof is Proposition $23$ of Book $\text{I}$ of Euclid's The Elements.
The careful reader will note that Proposition $22$: Construction of Triangle from Given Lengths does not directly create the necessary triangle at point $A$, but rather at a distance $CE$ from point $A$. However, a slight modification of the construction produces the desired triangle at the desired location.

This theorem is one of the two attributed to Oenopides of Chios (the other being Proposition $12$ of Book $\text{I} $: Perpendicular through Given Point).

Sources

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