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In AD bot BC, prove that AB^(2)+CD^(2)=B...

In `AD bot BC`, prove that `AB^(2)+CD^(2)=BD^(2)+AC^(2)`

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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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