In an isosceles triangle ABACwith AB=AC BD is perpendicular from B to the side AC. Prove that BD2-CD2= 2CD.AD
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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
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
sakshi8918:
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Step-by-step explanation:
Here is the req answer
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