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Iranian Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Iran

In the triangle $ABC$ the point $M$ is the midpoint of $AB$ , and the point $B'$ is the foot of the altitude from $B$ to $AC$ . The circle ( $CB'M$ ) intersects $BC$ again at $D$ . The circles ( $ABD$ ) and ( $CB'M$ ) intersect again at $K$ . The line parallel to $AB$ passing through $C$ intersects circle ( $CB'M$ ) again at $L$ . Prove that $KL$ bisects the segment $CM$ .

Solution

Since $CL \parallel AB$ and $A$ , $B$ , $D$ , $K$ are concyclic, we have
$180^\circ - \angle AKD = \angle ABD = \angle BCL = \angle DKL$
Thus, $\angle AKD + \angle DKL = 180^\circ$ . This implies that $A$ , $K$ , $L$ are collinear.

Notice that $B'M$ is the median to the hypotenuse and $CB'ML$ is cyclic, we have
$\angle MLC = \angle MB'A = \angle MAB'$
the quadrilateral $AMLC$ is a parallelogram and we can deduce the problem. ■

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