RMO 2011 Problem 1 | Angles of a triangle

This is a problem from Regional Mathematics Olympiad, RMO 2011 Problem 1 based on the angles of a triangle. Try it out!

Problem: RMO 2011 Problem 1

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 if 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 $

Some Useful Links:

• Our Math Olympiad Program
• AM-GM Inequality – Video