**Problem 2 of the British Mathematical Olympiad 2005**

In triangle ABC, ∠BAC = 120°. Let the angle bisectors of angles A, B and C meet the opposite sides in D, E and F respectively. Prove that the circle on diameter EF passes through D.

**Solution by Steve Dinh, a.k.a. Võ Đức Diên (dedicated to my former professor Le Chi De)**

Let BE meet CF at I, the incenter id ΔABC. Let J and S be the points of tangency of the inccircle with the sides BC and AC, respectively.

Link JS to meet BE at L. From E draw the perpendicular to CF to meet CF and BC at K and N, respectively. Also let CF meet ED at T, BE meet FD at V.

We have ∠BID = ∠ABI + ∠BAI = 90° - ½∠C = ∠JIC or ∠BIJ = ∠DIC

It’s easily seen that ∠EIK = ½ (∠B +∠C) = 30° or ∠BIC = 150° and ∠IEK = 90° - ∠EIK = 60° and since CI is the perpendicular bisector of EN, IE = IN and ∠INE = 60°. It follows that IEN is an equilateral triangle and ∠NIK = 30°.

We now have

∠IND = ∠NIK + ½∠C = 30° + ½∠C = 30° + (30° - ½∠B) = 60° - ½∠B and ∠DIN = ∠DIC - 30° = ∠BIJ - 30° = 90° - ½∠B - 30° = 60° - ½∠B

or ∠IND = ∠DIN and DE is bisector of ∠IEN or ∠DEN = 30°

From here, ∠ITE = 30° + 90° = 120°

Similarly on the other side, we have ∠IVF = 120°

But ∠ITE + ∠IVF = ∠BIC + ∠FDE or ∠FDE = ∠ITE + ∠IVF - ∠BIC = 90°

Therefore, the circle on diameter EF passes through D.

**Note**: The problem has been previously offered at the 56 Leningrad Mathematical Olympiad (1990). A different solution is available elsewhere.