# Prove AB^2+AC^2=2BM^2+2AM^2 for the diagram at http://i.imgur.com/hOYs3.jpg

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To prove the identity: AB^2+AC^2=2BM^2+2AM^2, we have to use the law of cosines.

For the triangle AMB,

c^2 = x^2 + y^2 - 2*x*y*cos (AMB)...(1)

For the triangle AMC

b^2 = x^2 + y^2 - 2*x*y*cos (AMC)

as cos (AMC) = cos( 180 - AMB) = -cos (AMB), we get

b^2 = x^2 + y^2 + 2*x*y*cos (AMB...(2)

Adding (1) and (2)

c^2 + b^2 = 2x^2 + 2y^2

c = AB , b = AC, x = BM, y = AM

=> AB^2 + AC^2 = 2BM^2 + 2AM^2

Therefore we prove **AB^2 + AC^2 = 2BM^2 + 2AM^2**

In the triangle ABC, M is the mid point of BC.

Draw a perpendicular from AD to BC, D being the feet of the perpendicular.

So in the figure BD = BM+DM. and CM = CD-DM.

Angle ADB = angle ADC = 90 deg.

So AB^2 = = BD^2+AD^2 = (BM+MD)^2.

AB^2 = (BM+MD)^2+AD^2 .....(1)

AC^2 = (CM-MD)^2+AD^2......(2)

(1)+(2): AB^2+AC^2 = 2BM^2+2MD^2+2AD^2, since CM = BM by data. And (x+y)^2+(x-y)^2 = 2x^2+2y^2, so (BM+MD)^2+(CM-MD)^2 = 2BM^2+2MD^2.

AB^2+AD^2 = 2BM^2 +2(BM^2+AD^2)

AB^2+AC^2 = 2BM^2+2AM^2, as the triangle ADM is right angled at D, MD^2+AD^2 = AM ^2.

**So AB^2+AC^2 = 2BM^2+2AM^2.**

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