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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
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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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