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Croatian Mathematical Competitions

Geometry Difficulty 5.4 AIME, harder Prove it Croatia

Let $ABCD$ be a cyclic quadrilateral such that $|AD| = |BD|$ and let $M$ be the intersection of its diagonals. Let $N$ be the second intersection of the diagonal $AC$ with the circle passing through $B$ , $M$ and the incentre of the triangle $BCM$ . Prove that $|AN| \cdot |NC| = |CD| \cdot |BN|$ .

Solution

We denote $\angle BAD = \angle DBA = \alpha$ in the isosceles triangle $ABD$ . Then $\angle ACB = \angle ADB = 180^\circ - 2\alpha$ .
Let $I$ be the incentre of the triangle $BCM$ . We have
$\begin{aligned} \angle MIB &= 180^\circ - \frac{1}{2} (\angle BMC + \angle CBM) = 180^\circ - \frac{1}{2} (180^\circ - \angle MCB) \\ &= 180^\circ - \frac{1}{2} (180^\circ - \angle ACB) = 180^\circ - \alpha. \end{aligned}$
Since the quadrilateral $BIMN$ is cyclic we can prove that $\angle BNC = \alpha$ (independently of the position of the point $N$ ) so the triangle $BCN$ is isosceles and $|BC| = |NC|$ .

Note that the triangles $ABN$ and $DBC$ are similar because $\angle ANB = \angle DCB = 180^\circ - \alpha$ and $\angle BAN = \angle BDC$ (suspended angles over the chord $BC$ ). Hence
$\frac{|NA|}{|BN|} = \frac{|CD|}{|BC|} = \frac{|CD|}{|NC|},$
i.e. $|AN| \cdot |NC| = |CD| \cdot |BN|$ .

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