Q13 of 97 Page 187

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.


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