Question

in ∆ABC, ∠C is an acute angle, seg AD Iseg 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: 

∴ LetAB = 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] 

∴ c² = (a – x)² + [P² ] (i) 

∴ c² = a² – 2ax + x² + [P² ] 

In ∆ADC, ∠D = 90° [Given] 

AC² = AD² + CD² [Pythagoras theorem] 

∴ b² = p² + [X² ] 

∴ p² = b² – [X² ] (ii) 

∴ c² = a² – 2ax + x² + b² – x² [Substituting (ii) in (i)] 

∴ c² = a² + b² – 2ax 

∴ AB² = BC² + AC² – 2 BC × DC