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:

Regional Math Olympiad 2011

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