Question

ans this que if you can

Answer 1

The question is wrong since there is no constraint on point D, hence q can be varied, implying that there is no relation among p,q and r.
Considering this as a misprint, here is the solution:

$\mathrm{By \: Pythagoras \: theorem,}\\ 2\sqrt{r^2-q^2}=\sqrt{r^2-p^2}\\ \text{On \: squaring \: and \: solving,}\\ \text{ \: we \: get \: the \: answer}$

Answer 2

let ac=a then ab=2a

seg om⊥ac seg on ⊥ seg ab

an=nb=a

am=mc=a

in ΔOMA

OA²=OM²+AM²

     =q²+a²/2²

       =q²+a²/4⇒1

in Δ ONA

OA² =AN²+ON²

      =a²+p² ⇒2

from eq 1 and 2

a²/4 + q²=p²+a²

a²+ 4q²=4p²+4a²

4q²= 4p²+3a²

4q²=p²+3(p²+a²)

4q²=p²+3r²                                                              from eq 2

hence proved