**Problem 1 of Asian Pacific Mathematical Olympiad 1993**

Let ABCD be a quadrilateral such that all sides have equal length and angle ABC is 60°. Let l be a line passing through D and not intersecting the quadrilateral (except at D). Let E and F be the points of intersection of l with AB and BC respectively. Let M be the point of intersection of CE and AF. Prove that CAČ = CM × CE.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to thi si MaiLy)**

Let a be the side length of the equilateral triangles ABC and ACD as shown.

and let ∠AEC = α and ∠ACE = β.

We have α + β = 180° - ∠EAC = 180° - 120° = 60°. (1)

We also have AE || CD and AD || CF; therefore, the two triangles EAD and DCF are similar, and we have

EA / a = a / CF.

This makes the two triangles EAC and ACF also similar, implying

α = ∠CAF.

Therefore from (1) ∠CAM + β = 60° and ∠AMC = 120°.

From here, the two triangles EAC and AMC are similar because their respective angles equal.

Therefore, CE/CA = CA/CM or CAČ = CM × CE

**The following problem is derived from the above problem:**

Let ABCD be a quadrilateral such that all sides have equal length and angle ABC is 60°. Let l be a line passing through D and not intersecting the quadrilateral (except at D). Let E and F be the points of intersection of l with AB and BC respectively. Let M be the point of intersection of CE and AF. Find the locus of point M.

The configuration has additional features which are illustrated dynamically elsewhere.