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

1.	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» 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]

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»

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]