Question

In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. 

Prove that: AB² = BC² + AC² + 2 BC × CD.

Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.

To prove: AB² = BC² + AC² + 2BC × CD

Answer 1

Proof:

Let AD = p, AC = b, AB = c, 

BC = a, DC = x 

BD = BC + DC [B – C – D] 

∴ BD = a + x 

In ∆ADB, ∠D = 90° [Given]

AB² = BD² + AD^{​​​​​​​2} [Pythagoras theorem]