**Problem 1 of the China Mathematical Olympiad 2010**

Circle Γ_{1} and Γ_{2} intersect at two points A and B. A line through B intersects Γ_{1} and Γ_{2} at points C and D, respectively. Another line through B intersects Γ_{1} and Γ_{2} at points E and F, respectively. Line CF intersects Γ_{1} and Γ_{2} at points P and Q, respectively. Let M and N be the midpoints of arcs PB and QB, respectively. Prove that if CD = EF, then C, F, M and N are concyclic.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Mimi Nguyen of IBM)**

Extend AB to meet CF at G. We are going to prove that BG is the bisector of ∠CBF. We have CB×CD = CQ×CF and FB× EF = FP×CF. Now dividing the two equations and taking into account CD = EF, we get CB/FB = CQ/FP . We also have

PG × CG = GB × GA = QG × FG or QG/PG = CG/FG = (QG + CG)/(PG + FG) = CQ/PF.

It follows that CG/FG = CB/FB or ∠CBG = ∠FBG and BG is the bisector of ∠CBF.

So now the three bisectors CM, FN and BG concur. Let them meet at I on BG. We now have IM × IC = IB × IA = IN × IF or IM/IN = IF/IC, meaning that the two triangles IMN and IFC are similar which implies ∠IMN = ∠IFC. On the other hand,

∠IMN + ∠CMN = 180° or ∠CMN + ∠NFC = 180° and C, F, M and N are concyclic.

**Extension of the problem**

Let’s prove that MN || O_{1}O_{2} where O_{1} and O_{2} are centers of Γ_{1} and Γ_{2}, respectively. Since ∠CBG = ∠FBG, we have ∠ABD = ∠FBG, i.e., AD = AF. Let K be the midpoint of arc FD, AK is then the diameter of Γ_{2}.

∠MCP = ½ ∠BCP = ½ arc (DF – BQ) = arc (KF – NQ) or ∠MCP + ∠NFQ = arc FK. Extend MN to meet Γ_{2} at L, ∠LNF = ∠MCP; therefore, ∠KNL = ∠NFQ = ∠BAN (subtending arc NB = NQ). But ∠ANK = 90° or AN ┴ NK; therefore, NL ┴ AG or MN || O_{1}O_{2}.