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Geometry Difficulty 4.4 AIME Prove it Ireland

In a triangle $ABC$ the internal bisector of the angle $\angle ABC$ meets the external bisector of $\angle BCA$ at $D$ . The circumcircle of the triangle $ABC$ cuts $BD$ at $E$ . Prove that $E$ is the circumcentre of the triangle $ACD$ .

Solution

Because $BD$ bisects $\angle ABC$ , we have $|EA| = |EC|$ .

Let $I$ be the incentre of $\triangle ABC$ , then $\angle DCI = 90^\circ$ .

We have $\angle EIC = \angle ICB + \angle IBC = \angle ACI + \angle ABE = \angle ACI + \angle ACE = \angle ECI$ .

Thus $\angle ECD = \angle DCI - \angle ECI = 90^\circ - \angle EIC = \angle EDC$ , which implies $|EC| = |ED|$ .

Therefore, $|EA| = |EC| = |ED|$ and $E$ is the circumcentre of $\triangle ACD$ .

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