Problem 4 of Asian Pacific Mathematical Olympiad 1998

Let ABC be a triangle and D the foot of the altitude from A. Let E and F be on a line through D such that AE is perpendicular to BE, AF is perpendicular to CF, and E and F are different from D. Let M and N be the midpoints of the line segments BC and EF, respectively. Prove that AN is perpendicular to NM.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Let w1 and w2 be the circles with diameters AB and AC, respectively.

E is on w1 because AE is perpendicular to BE.

F is on w2 because AF is perpendicular to CF.

Let AN intercept w1 at Q and w2 at P. Point N is outside w1, and we have

NE × ND = NQ × NA since N is also the midpoint of EF, we have

(1)NF × ND = NQ × NA

For w2, we have

(2)NF × ND = NP × NA

From (1) and (2), NQ × NA = NP × NA

or NQ = NP

Combine with MB = MC, we have MN || PC

Since AP is perpendicular to PC and MN || PC, AN is perpendicular to NM.