Question

In figure, angle B of triangle ABC is an acute angle and AD perpendicular to BC. Prove that AC²=AB²+BC²-2BC.BC​

Answer 1

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Given: A AABC in which zB is an acute

angle and AD 1 BC.

To Prove: A * C ^ 2 = A * B ^ 2 + B * C ^ 2 - 2BCBD .

Proof: Since AADB is a right triangle right-angled at D. So, by Pythagoras

theorem, we have

AB^ 2 =AD^ 2 +BD^ 2 .(i)

Again AADC is a right triangle right angled at D.

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So, by Pythagoras theorem, we have

A * C ^ 2 = A * D ^ 2 + D * C ^ 2

AC² = AD2 + (BC – BD)²

AC² = AD2 + (BC2 + BD2 - 2BC BD)

AC² = (AD² + BD2 ) + BC2 - 2BC BD

AC² = AB2+ BC2 - 2BC BD [Using

Hence, A * C ^ 2 = A * B ^ 2 + B * C ^ 2 - 2BCBD

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Answer 2

I

Given angleB=acute angle AD is perpendicular upon Bc.toFind proving (Ac)2=(AB)2+(BC)2