Question

In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC. 

Answer 1

In ΔABC, ray BD is the bisector of ∠ABC

∴ by the theorem of an angle bisector of a triangle,

`"AB"/"BC"="AD"/"DC"`      ...(1)

In ΔABC, ray CE is the bisector of ∠ACB

∴ by the theorem of an angle bisector of a triangle,

`"AC"/"BC"="AE"/"EB"`       ...(2)

Seg AB ≅ seg AC (Given)   ...(3) 

[From (1), (2) and (3)] 

∴ `"AB"/"BC"="AC"/"BC"`   ...(4)

In ΔABC, 

`"AE"/"EB"="AD"/"DC"`  ...[From (1), (2) and (4)]

∴ by converse of basic proportionality theorem,

seg ED || side BC

∴ ED II BC