Question

Let ABC be a triangle. Let D, E, F be points
respectively on segments BC, CA, AB such that
AD, BE, CF concurrent at the point K. Suppose
BD
BF and ADB = ZAFC. Prove that ZABE
and ZADB = ZAFC. Prove that ZABE =
Br
ZCAD.

Answer 1

Consider

∠BFC + ∠AFC = 180° (linear pair)

But ∠ADB = ∠AFC

=> ∠BFC + ∠ADB  = 180°

Thus, points B, F, K and D are concyclic (as opposite angles are supplementary)

i.e., quadrilateral BFKD is cyclic.

So, we can have a circle passing through all these points.

Now, ∠ABE = ∠FDK

(angles in a same segment are equal) ...(1)

It is given that  

⇒ FD || AC (by Basic proportionality Theorem)

Consider FD || AC and AD is the transversal.

∴ ∠FDA = ∠CAD

∠FDK = ∠CAD         ..............(2)

From (1) and (2), ∠ABE = ∠CAD.