In an isosceles triangle ABC, with AB=AC, BD is a perpendicular from B to the side AC. Prove that BD^2-CD^2=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
Read more on Brainly.in - https://brainly.in/question/702211#readmore
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
Read more on Brainly.in - https://brainly.in/question/702211#readmore
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