**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 w_{1} and w_{2} be the circles with diameters AB and AC, respectively.

E is on w_{1} because AE is perpendicular to BE.

F is on w_{2} because AF is perpendicular to CF.

Let AN intercept w_{1} at Q and w_{2} at P. Point N is outside w_{1}, and we have

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

(1) | NF × ND = NQ × NA |

For w_{2}, 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.