Question

In an isosceles triangle ABC with AB=AC, BD is perpendicular from B to the side AC. Prove that BD2-CD2=2CD.AD

Answer 1

Using Pythagoras theorem:

   AD² + BD² = AC²   (= AB²)       --- (1)
  CD² + BD² = BC²                 ------- (2)

(2) -  2 * (1)  =>
=>  CD² - BD² =  BC² - 2 AC² + 2 AD²
=>                   =  BC² - 2 (AD +CD)² + 2 AD² 
=>                   =  BC² - 2 CD² - 4 AD * CD
=>                   =  BD² + CD²  - 2 CD² -4 AD * CD
=>                   = BD² - CD² - 4 AD * CD
Simplify to get 
     BD² - CD² = 2 AD * CD

Answer 2

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Refer to the attached file for the answer...

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