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

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Sum

In ∆ABC, seg AD ⊥ seg BC, DB = 3CD.

Prove that: 2AB= 2AC+ BC2

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Solution

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

From (3) & (4)

2AB2 - 2AC2 = BC2

∴ 2AB2 = 2AC2 + BC2

  Is there an error in this question or solution?
Chapter 2: Pythagoras Theorem - Problem Set 2 [Page 45]

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Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 2 Pythagoras Theorem
Problem Set 2 | Q 13 | Page 45

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