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Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.
Ans: To prove AB2 + AC2 = 2AD2 + 2 (1/2 BC)2
Draw AE ⊥ BC
In Δ ABD since ∠D > 900
∴AB2 = AD2 + BD2 + 2BD x DE ....(1)
Δ ACD since ∠ D < 90o
AC2 = AD2 + DC2 - 2DC x DE ....(2)
Adding (1) & (2)
AB2 + AC2 = 2(AD2 + BD2)
= 2(AD2 + ( 1/2 BC )2)
Or AB2 + AC2 = 2 (AD2 + BD2)
Hence proved
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