Question

In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.

Prove that: 2AB² = 2AC² + BC²

Answer 1

In ∆ABC,  AD ⊥ BC, and BD = 3CD   ...(Given)                   

In ∆ADC, ∠ADC = 90°

By Pythagoras’ theorem,

AC² = AD² + CD² 

2AC² = 2AD² − 2CD²   ...(1) [Multiplied by 2]

In ∆ADB, ∠ADB = 90°

by Pythagoras’ theorem,

AB² = AD² + BD² 

2AB² = 2AD² + 2BD²   ...(2) [Multiplied by 2]

Subtracting equation (1) from (2)

2AB² − AC² = (2AD² + 2BD²) − (2AD² + 2CD²)   

= 2AD² + 2BD² − 2AD² − 2CD²

= 2(3CD)² − 2CD²   

= 2 × 9CD² − 2CD²  ...[Given] 

= 18CD² − 2CD²

= 16CD²

∴ 2AB² − 2AC² = 16CD²    ...(3)

BC = CD + DB    [C−D−B]

BC = CD + 3CD = 4CD

BC² = 16 CD²    ...(4) [squaring both sides]

From (3) & (4)

2AB² − 2AC² = BC²

∴ 2AB² = 2AC² + BC²