Question

ABC is an acute angled triangle and AD perpendicular BC. Prove that AC2=AB2-2BC.DB.​

Answer 1

Answer:

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