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Question
AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.
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Solution
In equilateral triangle AD ⊥ BC.
⇒ BD = DC = `"BC"/(2)` ...(In equilateral triangle altitude bisects the opposite side)
In right triangle ABD,
AB2 = AD2 + BD2
= `"AD"^2 + ("BC"/2)^2`
= `(4"AD"^2 + "BC"^2)/(4)`
= `(4"AD"^2 + "BC"^2)/(4)` ...(Since AB = BC)
⇒ 4AB2 = 4AD2 + AB2
⇒ 3AB2 = 4AD2
Hence proved..
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