Question

In triangle abc angle b = angle c, d and e are points on the sides ab and ac respectively such that bd = ec prove de parallel to bc

Answer 1

https://qph.ec.quoracdn.net/main-qimg-2988549ead50a14b6935aeaae5b59ba0-c
There is pic of the hint in this link
As shown in the above figure, in triangle ABC, D is on AC. E is on AB.

& AD = DE = EC = BC

TO PROVE:
Since AD = DE

=> < A = < AED = x ( isosceles triangle)…….(1)

=> < CDE = x + x = 2x ( exterior angle = the sum of 2 opposite interior angles)

But, DE =CE ( given)

So, < CDE = < DCE = 2x

=> < BEC = 2x + x = 3x ( exterior angle of triangle ACE )

And CE = CB ( given)

=> < CEB = < CBE = 3x …………(2)

By (1) & (2)

< A /