In the given figure, ΔABC is an obtuse triangle, obtuse-angled at B. If AD ⊥ CB (produced) prove that AC² = AB² + BC² + 2BC • BD.

Given: AD ⊥ CB (produced)

To prove: AC² = AB² + BC² + 2BC • BD

In ∆ ADC, DC = DB + BC ……….(i)

First, in ∆ ADB,

Using Pythagoras theorem, we have

AB² = AD² + DB²⇒ AD² = AB² – DB² ……….(ii)

Now, applying Pythagoras theorem in ∆ ADC, we have

AC² = AD² + DC²

= (AB² – DB²) + DC² [Using (ii)]

= AB² – DB² + (DB + BC)² [Using (i)]

Now, ∵ (a + b)² = a² + b² + 2ab

∴ AC² = AB² – DB² + DB² + BC² + 2DB • BC

⇒ AC² = AB² + BC² + 2BC • BD