Question

In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. 

Prove that: AB² = BC² + A² – 2 BC × DC.

Given: ∠C is an acute angle, seg AD ⊥ seg BC.

To prove: AB² = BC² + AC² – 2BC × DC

Answer 1

Proof:

∴ Let AB = c, AC = b, AD = p,

∴ BC = a, DC = x 

BD + DC = BC [B – D – C] 

∴ BD = BC – DC 

∴ BD = a – x 

In ∆ABD, ∠D = 90° [Given] 

AB² = BD² + AD² [Pythagoras theorem]