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In any triangle ABC, prove that AB^2 + A...

In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC.

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Let point B and C lie on the x-axis such that `BD=DC`, where, `D (0,0)` is midpoint of BC.
Let point A be (x,y).

`BD=k`.So, point B is `(-k,0)` and point C is `(k,0)`.
Now, `AB^2+AC^2=(x+k)^2+(y-0)^2+(x-k)^2+(y-0)^2`
`=x^2+k^2+2xk+y^2+x^2k^2-2xk+y^2`
`=2x^2+2y^2+2k^2=2(x^2+y^2+k^2)`
And `2(AD^2+BD^2)=2[(x-0)^2+(y-0)^2+k^2]`
`=2(x^2+y^2+k^2)`
Therefore, `AB^2+AC^2=2(AD^2+BD^2)`
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