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In a triangle A B C ,\ A C > A B , D ...

In a triangle `A B C ,\ A C > A B` , `D` is the mid-point of `B C` and `A E_|_B C` . Prove that: (i) `A B^2=A D^2-B C D E+1/4B C^2` (ii) `A B^2+A C^2=2\ A D^2+1/2B C^2`

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`triangleABC` in which AD is the median , i.e. BD= DC = `(BC)/2`
To prove : `AB^(2)+AC^(2)= 2AD^(2)+ 1/2BC^(2)`
Construction: Draw `AE bot BC`
Proof : L.H.S. = ` AB^(2) + AC^(2)`
`(AE^(2) + BE^(2)) + (AE^(2) + EC^(2))`
(by Pythagores theorem)
` 2AE^(2) + BE^(2) + EC^(2)`
`2(AD^(2)-DE^(2))+BD-ED)^(2))+ (ED+DC)^(2)`
`2AD^(2) - 2DE^(2)+BD^(2)+ED^(2)-2BD.ED+ED^(2)+DC^(2)+2ED . DC `
` = 2D^(2) + (1/2 BC)^(2) -2(BC)/2. ED+ (1/2BC)^(2) + 2ED(BC/2) " " (DB = DC = (BC)/(2)`
`= 2AD^(2) +1/2 BC^(2)`
R.H.S. Hence proved
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