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Given: In △ABC, ∠B = 90° and D is the mid-point of BC.
To Prove: AC2 = AD2 + 3CD2
Proof:
In △ABD,
AD2 = AB2 + BD2
AB2 = AD2 - BD2 .......(i)
In △ABC,
AC2 = AB2 + BC2
AB2 = AC2- BD2 ........(ii)
Equating (i) and (ii)
AD2 - BD2 = AC2 - BC2
AD2 - BD2 = AC2 - (BD + DC)2
AD2 - BD2 = AC2 - BD2- DC2- 2BDx DC
AD2 = AC2 - DC2 - 2DC2 (DC = BD)
AD2 = AC2 - 3DC2
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