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

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Draw AE⊥BC, 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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