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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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