TheoremProved

Inscribed angle theorem

Statement

Let $A$ , $B$ , $C$ be points on a circle with center $O$ . The inscribed angle $\angle ACB$ equals half the central angle $\angle AOB$ subtending the same arc $AB$ : $\angle ACB = \tfrac{1}{2}\angle AOB$ . Consequently, all inscribed angles subtending the same arc are equal, and any angle inscribed in a semicircle is a right angle ( $90^\circ$ ).

Why is it true?

Because every point on the circle lies at the same distance $R$ from the center $O$ , any chord from the vertex $C$ to $A$ or $B$ forms an isosceles triangle with two radii. The exterior angle at the center $O$ is the sum of the two equal base angles of that isosceles triangle, which is why looking at the arc $AB$ from the far rim of the circle always cuts the viewing angle at the center $O$ in half.

Proof sketch

First suppose the diameter $CD$ through $C$ and $O$ is one side of the inscribed angle, say $B = D$ . Then $\triangle OAC$ is isosceles with $OA = OC$ , so $\angle OCA = \angle OAC$ , and the exterior angle at $O$ satisfies $\angle AOB = \angle OCA + \angle OAC = 2\angle ACB$ . For the general case, draw the diameter $CD$ and either add or subtract the two diameter cases for arcs $AD$ and $BD$ to obtain $\angle ACB = \tfrac{1}{2}\angle AOB$ .

Proved by

proved -300Euclid of Alexandria

Topics that use this theorem

Circles

Related theorems

Law of sines Ptolemy's theorem Pythagorean theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Euclid (translated by Thomas L. Heath) (1956). The Thirteen Books of Euclid's Elements, Vol. 2 (Book III, Propositions 20–21, 31)
H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited · DOI:10.5948/UPO9780883859346