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In ∆ABC, if M is the midpoint of side BC and seg AM ⊥ seg BC, then prove that 

AB2 + AC2 = 2 AM​​​​​​​2 + 2 BM​​​​​​​2

Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC.

To prove: AB2 + AC2 = 2 AM2 + 2 BM2

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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)]

∴ AB + AC = 2 AM + BM + BM [∵ BM = MC (M is the midpoint of BC)]

∴ AB2 + AC2 = 2 AM2 + BM2 + BM2 [∵ BM = MC (M is the midpoint of BC)]

∴ AB2 + AC2 = 2 AM2 + 2 BM2

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