Proof:
In ∆AMB, ∠M = 90° [segAM ⊥ segBC]
∴ AB2 = AM2 + BM2 (i) [Pythagoras theorem]
Also, in ∆AMC, ∠M = 90° [seg AM ⊥ seg BC]
∴ AC2 = AM2 + MC2 (ii) [Pythagoras theorem]
∴ AB2 + AC2 = AM2 + BM2 + AM2 + MC2 [Adding (i) and (ii)]
∴ AB2 + AC2 = 2 AM2 + BM2 + BM2 [∵ BM = MC (M is the midpoint of BC)]
∴ AB2 + AC2 = 2 AM2 + 2 BM2