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Question
In triangle ABC, ∠B = 90o and D is the mid-point of BC.
Prove that: AC2 = AD2 + 3CD2.
Solution
In ΔABC
AB2 + BC2 = AC2
as [BC = 2CD]
AB2 + 4CD2 = AC2 ....(1)
In ΔABD,
AD2 = AB2 + BD2 = AB2 + CD2 ....(2)
Subtracting (1) and (2)
AC2 - AD2 = AB2 + 4CD2 - (AB2 + CD2)
AC2 - AD2 = 3CD2
AC2 = AD2 + 3CD2
Hence proved.
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