Track / Stage 7 / 218 of 300 #1618 of 2000

Problem 1618

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.5 Prove it Hong Kong Team Selection Test 2 · Hong Kong

Let $M$ be the midpoint of the side $BC$ of an acute $\triangle ABC$ , and let $D$ be the foot of perpendicular from $C$ to $AM$ . The circumcircle of $\triangle ABD$ intersects the side $BC$ again at $E \neq B$ . Suppose $F$ is a point on the segment $AE$ such that $FB = FC$ . Prove that $F$ is the midpoint of $AE$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let $P$ be the foot of perpendicular from $A$ to $BC$ . It follows from $\angle ADC = \angle APC = 90^\circ$ that $A$ , $D$ , $P$ , $C$ are concyclic. Considering the power of $M$ , we find that
$ME \times MB = MD \times MA = MP \times MC.$
Since $MB = MC$ , we have $ME = MP$ . Note that $FM$ is the perpendicular bisector of $BC$ . Therefore, $FM \parallel AP$ . As $ME = MP$ , it follows from the intercept theorem that $F$ is the midpoint of $AE$ .

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