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In the Following Figure, Ad is Perpendicular to Bc and D Divides Bc in the Ratio 1: 3. Prove that : 2ac2 = 2ab2 + Bc2

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Question

In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.

Prove that : 2AC2 = 2AB2 + BC2

Sum
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Solution

Here,
BD : DC = 1 : 3.
⇒ BD = `1/4"BC" and CD = 3/4`BC

AC2 = AD2 + CD2 and AB2 = AD2 + BD2 
Therefore,
AC2 - AB2 = CD2 - BD2

= `( 3/4"BC" )^2 - ( 1/4 "BC" )^2 `

= `9/16 "BC"^2 -  1/16 "BC"^2`

= `1/2"BC"^2`

∴ 2AC2 - 2AB2 = BC2
2AC2 = 2AB2 + BC2
Hence proved.

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Chapter 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [Page 164]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 15 | Page 164

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