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Mathematical Olympiad in Ukraine

Geometry Difficulty 4.9 AIME Prove it Ukraine

On the side $AB$ of acute-angled triangle $ABC$ there is a point $K$ , $M$ is the midpoint of $BC$ , segments $AM$ and $CK$ intersect at a point $F$ . It is known that $KF = AK$ . Prove that $CF = AB$ .

Solution

On the ray $AM$ take a point $Q$ , such that $MQ = AM$ (fig.19).

Then $ACQB$ is a parallelogram and $\angle KAF = \angle FQC = \angle CFQ \Rightarrow CF = CQ = AB$ , and we are done.

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