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# 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 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 ${\angle KBF}$ = $\angle DAC$