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Geometry Difficulty 5.3 AIME, harder Prove it Ireland

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

Solution

Let the bisector of ∠BAC\angle BAC meet the circumcircle of △ABC\triangle ABC at DD.
Figure 1
Because ∠BAD=∠DAC\angle BAD = \angle DAC, we have ∣DB∣=∣DC∣|DB| = |DC|. On the other hand, ∠BID=∠BAI+∠ABI=∠CAI+∠CBI=∠CBD+∠CBI=∠DBI\angle BID = \angle BAI + \angle ABI = \angle CAI + \angle CBI = \angle CBD + \angle CBI = \angle DBI and so ∣DB∣=∣DI∣|DB| = |DI|. This shows that ∣DB∣=∣DC∣=∣DI∣|DB| = |DC| = |DI|, thus DD is the circumcentre of △BIC\triangle BIC.

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