Correct Option: B

In ||gm ABCD , we have
Since diagonals of parallelogram(||gm ) bisect each other,
∴ M will be the mid-point of each of the diagonal AC and BD
∴ In ΔABC AB² + BC² = 2(AM² + MB²) ......... ( 1 ) [Appolonius Theorem]
In ΔADC AD² + CD² = 2(AM² + DM²) ......... ( 2 )
= 2(AM² + MB²) [∴ DM = BM]
Adding equations ( 1 ) and ( 2 )
AB² + BC² + CD² + DA² = 2(AM² + MB² + 2(AM² + MB² = 4AM² + 4MB²
= (2AM)² + (2MB)²= AC²+ BD². { ∴ AM = MC , MB = MD }