Question

In ∆ABC, AD is a median. Prove that AB² + AC² = 2AD² + 2DC².

Answer 1

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`