Question

In Fig., ∆ABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CB, prove that AC² = AB² + BC² + 2BC × BD

Answer 1

Given: An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.

To Prove:  AC² = AB² + BC² + 2BC × BD

Proof: Since ∆ADB is a right triangle right angled at D. Therefore, by

Pythagoras theorem, we have AB² = AD² + DB² ….(i)

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

Therefore, by Phythagoras theorem, we have

AC² = AD² + DC²

⇒ AC² = AD² + (DB + BC)²

⇒ AC² = AD² + DB² + BC² + 2BC • BD

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

Hence, AC² = AB² + BC² + 2BC • BD