sum of two sides multiplied on 2 opposite points

[photon]

in a cyclic quadrilateral $PQRS$, lines through $Q$ and $S$ parallel to diagonal $PR$ keep perpendicular difference $d$.if the radius of the circle inscribed quadrilateral is $R$,then prove that,
$PS.SR+PQ.QR=2R.d$
[hint:$d$=sum of perpendiculars to $PR$ from $Q$ and $S$]

[*Mahi*]

By applying sine law on $\bigtriangleup PSR$,
$ \frac {PR} {sin \angle PSR} =2R$
Now the area of quadrangle $PQRS$ is
$(PQRS)=(PSR)+(PQR)$
$=\frac {1} {2} .PS.SR.sin \angle PSR+ \frac {1} {2} PQ.QR. sin \angle PQR$
$=\frac {1} {2} sin \angle PSR(PS.SR+PQ.QR)$
$=\frac {1} {2} . \frac {PR} {2R} (PS.SR+PQ.QR)$
Again ,
$(PQRS)= \frac {PR.d} {2}$
So,$PR.d= \frac {PR} {2R} (PS.SR+PQ.QR)$
Or,$2R.d=(PS.SR+PQ.QR)$ [Proved]

[photon]

i found it with another way,brahmagupta theorem.
but,i don't understand,
\[(PQRS)=\frac{1}{2}PR.d\]
please,explain.
edit:i got it.let perpendicular from Q,S are QM,SN to PR.
\[PQRS=\frac{1}{2}PR.QM+\frac{1}{2}PR.SN=\frac{1}{2}PR.d\]
as $QM+SN=d$
your proof was cool but a little bit hard to get......

[*Mahi*]

Right you are.........I didn't got much time to explain it plainly.........