In ∆ABC, AD is a median.

Solution:

We have the following figure.

Since triangle ABM and ACM are right triangles right angled at M

$A B^{2}=A M^{2}+B M^{2} \ldots \ldots(i)$

$A C^{2}=A M^{2}+C M^{2} \ldots \ldots(i i)$

Adding (i) and (ii), we get

$A B^{2}+A C^{2}=2 A M^{2}+B M^{2}+C M^{2}$

Since in triangle ADM we have

$A D^{2}=D M^{2}+A M^{2}$

So,

$A B^{2}+A C^{2}=2\left(A D^{2}-D M^{2}\right)+B M^{2}+C M^{2}$

$=2 A D^{2}-2 D M^{2}+B M^{2}+C M^{2}$

$=2 A D^{2}-2 D M^{2}+B M^{2}+C M^{2}+2 B M \times C M-2 B M \times C M$

$=2 A D^{2}-2 D M^{2}+(B M+C M)^{2}-2 B M \times C M$

BM + CM = BC

So,

$A B^{2}+A C^{2}=2 A D^{2}-2 D M^{2}-2 M B \times C M+B C^{2}$

Now we have

$B C^{2}=(2 C D)^{2}=4 C D^{2}$

So,

$A B^{2}+A C^{2}=2 A D^{2}-2 D M^{2}-2 M B \times C M+4 C D^{2}$

$A B^{2}+A C^{2}=2 A D^{2}-2 D M^{2}-2 M B \times C M+4 C D^{2}$

$=2 A D^{2}+4 C D^{2}-2 D M^{2}-2(C D+D M)(C D-D M)$

$=2 A D^{2}+4 C D^{2}-2 D M^{2}-2 C D^{2}+2 C D \times D M-2 D M \times C D+2 D M^{2}$

$=2 A D^{2}+2 C D^{2}$

Hence proved $A B^{2}+A C^{2}=2 A D^{2}+2 C D^{2}$