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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.4 AIME Prove it Saudi Arabia

In a triangle $ABC$ , let $K$ be a point on the median $BM$ such that $CM = CK$ . It turned out that $\angle CBM = 2\angle ABM$ . Show that $BC = KM$ .

Solution

Let $L$ be the reflection of $C$ through $BM$ . Since $CK = CM$ , we have $\angle CKM = \angle CMK$ . On the other hand, since $L$ is the reflection of $C$ through $KL$ , we have $\angle LKM = \angle CKM = \angle CMK$ , implying that $KL$ is parallel to $CM$ .

Also, we have $KL = KC = CM = MK$ , and $KL$ is parallel to $AM$ . Hence, $KMAL$ is a parallelogram, implying that $KM = AL$ and $KM$ is parallel to $AL$ . We have $\angle LBM = \angle CBM = 2\angle ABM$ , hence $\angle KBA = \angle ABL$ . Since $AL$ is parallel to $MB$ , we have $\angle BAL = \angle MBA = \angle LBA$ , implying that $LB = LA$ . Hence, we have
$KM = LA = LB = BC$
as desired.
$\square$

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