In `DeltaABC`,ray BD bisects `angleABC` and ray CE bisects `angleACB`. If seg AB `cong`seg AC, then prove that ED `abs()` BC.

In DeltaABC,"seg " BD" bisects "angleABC and"Ray " CE" bisects " angleACB. "if seg "AB~="seg "AC then prove that ED||BC.

In Delta ABC , ray BD bisects angleABC . A-D-C, side DE abs() side BC, A-E-B, then prove that (AB)/(BC) = (AE)/(EB) .

In Delta ABC , ray BD bisects angleABC . A-D-C , side DEabs()side BC , A-E-B , then prove that (AB)/(BC) = (AE)/(EB) .

In DeltaABC, ray BD bisects angleABC,A-D-C, Side DE|| Side BC. A-E-B. then prove that AB:BC=AE:EB

In DeltaABC ray BD bisects angleABC " "A-D-C,"side DE || side BC" A-E-B "then prove",(AB)/(BC)=(AE)/(EB)

In the figure, seg AB~= seg AC , ray CE bisects /_ACB , ray BD bisects /_ABC . Prove that ray ED|| side BC .

In quadrilateral ABCD, AB=AD, AC bisects angleA . Prove DeltaABC~ADC .

In the given figure, DeltaABC is an isosceles triangle in which AB=AC and AE bisects angleCAD . Prove that AE||BC .

In DeltaABC, AB = AC and the bisectors of angleB and angleC meet AC and AB at point D and E respectively. Prove that BD = CE .

In DeltaABC, AB = AC and the bisectors of angleB and angleC meet AC and AB at point D and E respectively. Prove that BD = CE .