# RMO 2011 Problem 1

Problem:

Let ABC be a triangle. Let D, E, F be points respectively on the segments BC, CA, AB such that AD, BE, CF concur at the point K. Suppose BD/DC = BF/FA and $$\angle ADB = \angle AFC$$. Prove that $$\angle ABE = \angle CAD$$.

Solution:

FKBD is a cyclic quadrilateral since opposite exterior angle $$(\angle CFA )$$ is equivalent to its interior opposite angle $$(\angle BDA )$$. Since FKBD is cyclic, its vertices lie on a circle and therefore FK is a segment of the circle. Angles on the same side of a segment of a circle are equal. Therefore, $${\angle KBF}$$ = $$\angle FDK$$ .

So, we need to prove that $${\angle FDK}$$ = $$\angle DAC$$

In a triangle say PQR, let D and E be points on QP and QR respectively. DE is parallel to PR iff QD/DP=QE/ER.In triangle ABC, BD/DC=BF/FA

Therefore FD is parallel to AC. Since angle FDK and angle DAC are alternative interior opposite angles when FD is parallel to AC and AD is the transversal.

Therefore $${\angle KBF}$$ = $$\angle FDK$$ = $$\angle DAC$$.

Hence proved $${\angle KBF}$$ = $$\angle DAC$$