In an isosceles triangle abc with ab is equal to ac bd is perpendicular from b to side bc then prove that bd square minus b square is equal to 2 into cd into a d
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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
Hope it helped........
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