Question

In an iscosceles triangle ABC, with AB=AC, BD is perpendicular to side AC. Prove that BD²-CD²= 2CDxAD

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 CD×AD

Hope it helped....