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. V Đức Din (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 - ∠C = ∠JIC or ∠BIJ = ∠DIC

Its easily seen that ∠EIK = (∠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 + ∠C = 30 + ∠C = 30 + (30 - ∠B) = 60 - ∠B and ∠DIN = ∠DIC - 30 = ∠BIJ - 30 = 90 - ∠B - 30 = 60 - ∠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.