In `DeltaABC,` if M is the midpiont of BC and seg AM `bot` seg BC, then prove that prove that `AB^(2)+AC^(2)=2AM^(2)+2BM^(2).`

Explanation

Proof `:` Let's consider the case, where AM is not perpendicular to seb BC, then out of `/_AMB ` and `/_ AMC`, one is acute angle and other is obtuse angle.
Let's consider `/_AMB ` as an acute angle and `/_AMC ` as an obtus angle.
`:. AMB ` is an acute angled triangle,
`:.` by application of Pythagoras theorem, we get,
`:. AB^(2) = AM^(2) + BM^(2) - 2BM.DM` ...(1)
`Delta AMC ` is an obtuse angled triangle.
By application of Pythagoras theorem, we get ,
`AC^(2) = AM^(2) + MC^(2) + 2MC.DM` ...(2)
But MC= BM ...(M is the midpoint of side BC ) ...(3)
Substituting (3) in (2), we get
`AC^(2) = AM^(2) + BM^(2) + 2BM.DM ` ..(4)
Adding (1) and (4) , we get ,
`AB^(2) + AC^(2) = AM^(2) + BM^(2) - 2BM.DM+ AM^(2) + BM^(2) + 2BM.DM`
`:. AB^(2) + AC^(2) = 2AM^(2) + 2BM^(2)`
Let's consider the case, where seg `AM _|_` seg BC
In `Delta AMB`,
`/_ AMB = 90^(@)`
`:.` by Pythagoras theorem,.
`AB^(2) = AM^(2) + BM^(2)` ...(5)
In `Delta AMC `,
`/_ AMC = 90^(@)`
`:. ` by Pythagoras theorem,
`AC^(2) = AM^(2) + MC^(2) `
But MC = BM .....(M is the midpoint of side BC ) ...(7)
`:. AC^(2) = AM^(2) + BM^(2)` ....[From (6) and ( 7) ] ....(8)
Adding (5) and (8), we get,
`AB^(2) + AC^(2) =AM^(2) + BM^(2) + AM^(2)+ BM^(2) `
`:. AB^(2) + AC^(2) = 2AM^(2) + 2BM^(2)`