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Irish Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Ireland

Let $I$ be the incentre of $\triangle ABC$ . Prove that the circumcentre of $\triangle BIC$ lies on the circumcircle of $\triangle ABC$ .

Solution

Let the bisector of $\angle BAC$ meet the circumcircle of $\triangle ABC$ at $D$ .

Because $\angle BAD = \angle DAC$ , we have $|DB| = |DC|$ . On the other hand, $\angle BID = \angle BAI + \angle ABI = \angle CAI + \angle CBI = \angle CBD + \angle CBI = \angle DBI$ and so $|DB| = |DI|$ . This shows that $|DB| = |DC| = |DI|$ , thus $D$ is the circumcentre of $\triangle BIC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.