Question

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. Suppose \(\frac{BD}{DC}\) = \(\frac{BF}{FA}\) and ∠ADB = ∠AFC, then 

(a) ∠ABE = ∠CAD 

(b) ∠ABE = ∠AFC 

(c) ∠ABE = ∠FKB 

(d) ∠ABE = ∠BCF

Answer 1

(a) ∠ABE = ∠CAD 

Since \(\frac{BD}{DC}\) = \(\frac{BF}{FA}\)

By basic proportionality theorem, we have FD || AC. 

Also, ∠BDK = ∠ADB = ∠AFC = 180º – ∠BFK 

∴ ∠BDK + ∠BFK = 180º 

⇒ BDKF is a cyclic quadrilateral 

⇒ ∠FBK = ∠FDK                   (Angles in the same segment) 

∴ ∠ABE = ∠FBK = ∠FDK = ∠FDA = ∠DAC 

(∵ FD || AC, alt. ∠s are equal)