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Pythagoras Theorem

In A B C, A B...

Question

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

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Solution

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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