Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Prove it Belarus

1. The extension of the median AM of the triangle ABC intersects its circumcircle at D. The circumcircle of the triangle CMD intersects the line AC at C and E. The circumcircle of the triangle AME intersects the line AB at A and F.
Prove that CF is an altitude of the triangle ABC.

Solution

1. It is enough to prove the equalities $FM = BM = MC$ from which it will follow

that $M$ is the midpoint of the hypotenuse of the right triangle $CFB$ and in particular $\angle CFB = 90^\circ$ .

Since the quadrilateral $ABDC$ is cyclic, it follows that $\angle ABC = \angle ADC$ .
Since points $M$ , $D$ , $E$ and $C$ lie on the same circle, $\angle MDC = \angle MEC$ .
From the cyclic quadrilateral $AFME$ we obtain the equalities $\angle BFM = \angle MEA = \angle FBM$ .
Therefore the triangle $BFM$ is isosceles with $FM = BM$ and $FM = BM = MC$ .

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