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Irish Mathematical Olympiad

Geometry Difficulty 4.4 AIME Prove it Ireland

$P$ is a point on a diameter $AB$ of a circle, centre $O$ . The points $C$ and $D$ are on the circumference of the circle, on the same side of $AB$ , such that $\angle APC = \angle BPD$ . Prove that the quadrilateral $PODC$ is cyclic.

Solution

Extend $CP$ to meet the circumference at $E$ . Then, using the assumption, we get $\angle BPE = \angle APC = \angle DPB$ , hence $D$ is the reflection of $E$ in the line $AB$ . Therefore, $ED$ is perpendicular to $AB$ and so $\angle BPE + \angle PED = 90^\circ$ .

We also have $\angle COD = 2\angle CED = 2\angle PED$ , as these angles are subtended by the same arc $CD$ . Because $|OC| = |OD|$ the triangle $CDO$ is isosceles and we obtain $2\angle BPE = 180^\circ - 2\angle PED = 180^\circ - \angle COD = 2\angle DCO$ . This implies $\angle DPB = \angle BPE = \angle DCO$ , hence $PODC$ is cyclic.

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