Question

AD is median of triangle ABC. The bisector of angle ADB and angle ADC meet AB and AC in E and F respectively. Prove that EF is parallel to BC.
{please explain deeply}

Answer 1

In tri. DAE, DE Bisect angle ADB
So we have
DA/DB=AE/EB equ.1
Similarly in tri. DAC, DE Bisect angle ADC
we get
DA/DC=AF/FC
(DC=DB)
DA/DB=AF/FC equ.2
From equ 1 and equ 2
AE/EB= AF/FC
In tri ABC
EF ¶ BE (... BPT)