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Iranian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Iran

Let $\text{ABCD}$ be a cyclic quadrilateral with incircle $\omega$ . Let $M$ be the midpoint of arc $\text{ABC}$ . Let $\Gamma$ be the circle with center $M$ and radius $MA$ . $AD$ and $AB$ intersect $\Gamma$ at $X$ and $Y$ , respectively. Let $Z$ be a point on line $XY$ (with $Z \neq Y$ ) such that $BY = BZ$ . Prove that $\angle BZD = \angle BCD$ .

Solution

Since $M$ is the center of the circumcircle of $AYC$ and $ABMC$ is also cyclic, we have:
$\begin{align*} \angle ABC &= \angle AMC = 2 \times \angle AYC \\ &= \angle BYC + \angle BCY \\ &= \angle AYC + \angle BCY \end{align*}$
Thus $BY = BC$ . Also, using the cyclicity of $AXYC$ , we have:
$\begin{align*} \angle ZBC &= 2 \times \angle ZYC \\ &= \angle ZBD + \angle CBD \\ &= \angle ZBD + \angle CAD \\ &= \angle ZBD + \angle ZYC \end{align*}$
Thus $\angle ZBD = \angle ZYC = \angle CBD$ , which implies $\triangle ZBD \cong \triangle CBD$ , and the proof is complete.

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