Question

In ABCD side AB =side AD. Bisector of angle BAC cuts side BC at E and bisector of angle DAC cuts DC at F. Prove that seg EF II seg BD​

Answer 1

Step-by-step explanation:

here

AE is angel bisecter of angel BAC

so BE/EC=AB/Ac-------angle bisecter th. ---1

also AF is bisecter of angel DAC

so DF/FC=AD/AC-------2

and AB=AD-------3

fom eq 1,2and3

BE/EC=DF/FC

so the EF || BD--------by convers of Basic proportionlity th.