**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 \({\angle KBF}\) = \(\angle DAC \)

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