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प्रश्न
In ∆ABC, AD is a median. Prove that AB2 + AC2 = 2AD2 + 2DC2.
उत्तर
We have the following figure.
Since triangle ABM and ACM are right triangles right angled at M
`AB^2=AM^2+BM^2`.....(1)
`AC^2=AM^2+CM^2`.....(2)
Adding (i) and (ii), we get
`AB^2+AC^2=2AM^2+BM^2+CM^2`
Since in triangle ADM we have
`AD^2=DM^2+AM^2`
So,
`AB^2+AC^2=2(AD^2-DM^2)+BM^2+CM^2`
`=2AD^2-DM^2+BM^2+CM^2`
`=2AD^2-DM^2+BM^2+CM^2+2BMxxCM-2BMxxCM`
`= 2AD^2-2DM^2+(BM+CM)-2BMxxCM`
BM + CM = BC
So,
`AB^2+AC^2=A2AD^2-2DM^2-2MBxxCM+BC^2`
`AB^2+AC^2=2AD^2-2DM62-2MBxxCM+4CD^2`
`=2AD^2+4CD^2-2DM^2-2(CD+DM)(CD-DM)`
`=2AD^2+4CD^2-2DM^2-2CD^2+2CDxxDM-2DMxxCD+2DM^2`
`= 2AD^2+2CD^2`
Hence proved `AB^2+AC^2=2AD^2+2CD^2`
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