**Problem 1 of Asian Pacific Mathematical Olympiad 1991**

Let G be the centroid of triangle ABC and M be the midpoint of BC. Let X be on AB and Y on AC such that the points X, Y , and G are collinear and XY and BC are parallel. Suppose that XC and GB intersect at Q and Y B and GC intersect at P. Show that triangle MPQ is similar to triangle ABC.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (tặng bạn James Bon Nguyen)**

Let N and D be midpoints of AC and AB, respectively. Since XY || BC, we have: YC / YN = GB / GN

or (YC / YN) × (GN / GB) = 1

M is midpoint of BC; therefore, ( MB / MC ) × ( YC / YN ) × ( GN / GB ) = 1

Per Ceva’s theorem, the three lines MN, GC and BY are concurrent and meet at P.

G is the centroid of triangle ABC, link CD. Since MN || AB and D is midpoint of AB, P is then midpoint of MN. We have MP || AB.

Using the same argument, Q is midpoint of MD and MQ || AC.

Therefore PQ || DN

In addition with DN || BC, we have PQ || BC.

ΔMPQ has each of its side parallel to each of those of ΔABC; therefore, they are similar.