Hello, devi!
The problem has strange wording ... and the statement is not true.
In $\displaystyle \Delta ABC,\;\angle A = 90^o$ and D is the midpoint of AC.
The value of: .$\displaystyle BC^2-BD^2 \:= \:AD^2$ . . . . "the value of..." ?
Prove. Consider a 3-4-5 right triangle. Code:
B*
| * *
| * * 5
3| * *
| * *
| * *
* - - - - - - * - - - - - - - *
A 2 D 2 C
We have: .$\displaystyle BC = 5,\;BD = \sqrt{13},\;AD = 2$
. . But: .$\displaystyle (5)^2 -(\sqrt{13})^2 \;\;{\bf{\color{red}\neq}}\;\;(2)^2$