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Mongolian National Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Mongolia

Let $\omega$ be the circumcircle of a scalene triangle $ABC$ . The tangents to $\omega$ at $A$ and $C$ meet in $P$ , and the line $BP$ intersects $\omega$ in $D$ . Let $BB'$ be a diameter of $\omega$ . The exterior angle bisector of $\angle ABC$ and the lines $B'A$ and $B'C$ intersect in $A'$ and $C'$ , respectively. Prove that $A', B', C', D$ are cyclic.

Solution

Since $PC$ is a tangent to $\omega$ we have $\triangle PCD \cong \triangle PBC$ and so $CD \cdot PB = PC \cdot CB$ . Similarly, $AD \cdot PB = PA \cdot AB$ . Thus
$\frac{CD}{CB} = \frac{AD}{AB}. \qquad (*)$
since $PA = PC$ . We also have that $\triangle AA'B \sim \triangle CC'B$ because of $\angle ABA' = \angle CBC'$ and $\angle A'AB = \angle BCC' = 90^\circ$ . Hence $\frac{AB}{CB} = \frac{AA'}{CC'}$ . Thus it follows from (*) that $\frac{AA'}{CC'} = \frac{AD}{CD}$ . Hence $\triangle DAA' \sim \triangle DCC'$ since $\angle B'AD = \angle B'CD$ which follows from $\angle DAA' = \angle DCC'$ . Therefore $\angle AA'D = \angle B'C'D$ , implying that $B', A', C', D$ are cyclic.

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