AD is a median of △ABC. The bisector of ∠ADB and ∠ADC meet AB and AC in E and F respectively. Prove that EF||BC.
In △DAE, DE Bisect ∠ADB
So we have
DADB=AEEB ------1
Similarly in △DAC, DE Bisect ∠ADC
we get
DADC=AFFC --- (DC=DB)
DADB=AFFC ------2
From 1 and 2
=AEEB=AFFC
In △ABC,
EF∥BE (Baisc Proportionality Theorem)