Question

In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that AC² = AB² + BC² – 2BC × BD

Answer 1

Given: A ∆ABC in which ∠B is an acute angle and AD ⊥ BC.

To Prove: AC² = AB² + BC² – 2BC × BD.

Proof: Since ∆ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have

AB² = AD² + BD² ….(i)

Again ∆ADC is a right triangle right angled at D.

So, by Pythagoras theorem, we have

AC² = AD² + DC²

⇒ AC² = AD² + (BC – BD)²

⇒ AC² = AD² + (BC² + BD² – 2BC • BD)

⇒ AC² = (AD² + BD² ) + BC² – 2BC • BD

⇒ AC² = AB² + BC² – 2BC • BD [Using (i)]

Hence, AC² = AB² + BC² – 2BC • BD