We know that diagonals of a parallelogram bisect each other. So, B is the midpoint of SQ.
Here, PB is a median of Δ QPS and we know that, a median of a triangle divides it into two triangles of equal area.
∴ ar(Δ BPQ) = ar (ΔBPS) ...(i)
ar(Δ OBQ) = ar (ΔOBS) ....(ii)
On adding Eqs. (i) and (ii), we get,