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Question

In ∆ABC, AB = AC. Side BC is produced to D. Prove that
(AD2 − AC2)=BD⋅CD.

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Solution

Draw AEBC, meeting BC at D.
Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.
So, BE = CE
and DE+ CE = DE + BE = BD


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