Construction of Segment on Given Circle Admitting Given Angle

Theorem

In the words of Euclid:

From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle.

(The Elements: Book $\text{III}$: Proposition $34$)

Construction

Let $ABC$ be the given circle, and let $D$ be the given rectilineal angle.

Let $EF$ be drawn tangent to $ABC$ at $B$.

Let $\angle FBC$ be constructed equal to $\angle D$.

Then $BC$ is a straight line which forms the base of a segment which admits an angle equal to $D$.

Proof

Let $A$ be selected anywhere on the circle opposite the segment in question.

From Angles made by Chord with Tangent‎, $\angle BAC = \angle FBC$.

But $\angle FBC = \angle D$.

Hence the result.

$\blacksquare$

Historical Note

This proof is Proposition $34$ of Book $\text{III}$ of Euclid's The Elements.

Sources

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