Question

In an obtuse ΔABC (∠B is obtuse), AD is perpendicular to CB produced. Then 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, ΔADB is a right triangle, right-angled at D.

∴ Pythagoras theorem, we have

AB² = AD² + DB²   ...(i)

Again, ΔADC is a right triangle, right-angled at D.

∴ AC² = AD² + DC²   ...(By Pythagoras Theorem)

⇒ AC² = AD² + (DB + BC)²

⇒ AC² = AD² + DB² + BC² + 2DB × BC

⇒ AC² = AB² + BC² + 2DB × BC   ...[Using equation (i)]

Hence Proved.