Question

In ∆ABC, seg AD ⊥ seg BC and DB = 3 CD. Prove that: 2AB² = 2AC² + BC². 

Given: seg AD ⊥ seg BC 

DB = 3CD 

To prove: 2AB² = 2AC² + BC²

Answer 1

DB = 3CD (i) [Given] 

In ∆ADB, ∠ADB = 90° [Given] 

∴ AB² = AD² + DB² [Pythagoras theorem] 

∴ AB² = AD² + (3CD)² [From (i)] 

∴ AB² = AD² + 9CD² (ii) 

In ∆ADC, ∠ADC = 90° [Given] 

∴ AC² = AD² + CD² [Pythagoras theorem] 

∴ AD² = AC² – CD² (iii)

AB² = AC² – CD² + 9CD² [From (ii) and(iii)] 

∴ AB² = AC² + 8CD² (iv) 

CD + DB = BC [C – D – B] 

∴ CD + 3CD = BC [From (i)] 

∴ 4CD = BC 

∴ CD = (BC/4) (v) 

AB² = AC² + 8( BC/4)² [From (iv) and (v)] 

∴ AB² = AC² + 8 × BC²/16

∴ AB² = AC² + BC²/2

∴ 2AB² = 2AC² + BC² [Multiplying both sides by 2]