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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

In a circle $O$ , there are six points, $A, B, C, D, E, F$ in a counterclockwise order. $BD \perp CF$ , and $CF, BE, AD$ are concurrent. Let the perpendicular from $B$ to $AC$ be $M$ , and the perpendicular from $D$ to $CE$ be $N$ . Prove that $AE \parallel MN$ .

Solution

Let $K$ be the concurrent point of $CF, BE, AD$ , and let $CF$ intersect $BD$ at $L$ .

We have $\angle BMC = 90^\circ = \angle BLC$ since $BM$ is perpendicular to $AC$ and $BL$ is perpendicular to $CF$ , so $B, M, L, C$ lie on the same circle with diameter $BC$ . Hence,
$\angle CML = \angle CBL = \angle CBD = \angle CAD$
implying that $ML$ is parallel to $AK$ . Similarly, we also have $LN$ is parallel to $EK$ . Thus, we have
$\frac{CM}{CA} = \frac{CL}{CK} = \frac{CN}{CE}$
This implies that $MN$ is parallel to $AE$ , as desired. $\square$

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