Question

AD is a median of triangle ABC. The bisector of angle ADB and angle ADC meet AB and AC at E and F respectively. Prove that EF is parallel to BC.

Answer 1

AE/EB = AD/BD and AF/FC = AD/DC (angle bisector theorem) 

But BD=DC hence AE/EB = AF/FC 

∴ EF||BC (similar triangles)
Hope it helps

Answer 2

In ΔDAB, DE bisects ∠ADB

∴ DA/ DB = AE/EB -- (1)

in ΔDAC, DF bisects ∠ ADC  

∴ DA/DC = AF/FC

=> DA/DB = AF/ FC --- (2)

From (1) and (2) we will get  

AE/EB = AF/FC

=> In ΔABC

EF║BC