In the given figure, ABC is a triangle in which AB=AC and D is a point on AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD=AE.

Given that AB = AC and also DE || BC.
So by Basic proportionality theorem or Thales theorem,
= ![]()
∴
= ![]()
Now adding 1 on both sides,
+ 1 =
+ 1
= ![]()
=
… as AB = AD + DE and AC = AE + EC
But is given that AB = AC,
∴
= ![]()
Hence,
AD = AE.