Question

AD is the median of traingle ABC and bisector of angle ADB , DE and DF meet AB at E and AC at F . prove that EF || BC

Answer 1

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