For this question we can use the theorem that a line joining the mid points of 2 sides of the triangle is parallel to the third side and equal to half of its length.
P and Q are mid-points of the sides AB and BC of triangle ABC.
=> PQ is parallel to side AC of triangle ABC and of length = (1/2)AC.
R and S are mid-points of the sides CD and DA of triangle ACD
=> RS is parallel to side AC of triangle ACD and of length = (1/2)AC
=> PQ and RS which are the opposite sides of the quadrilateral PQRS are of equal length and both being parallel to AC are parallel to each other.
=> quadrilateral PQRS is a parallelogram.
Also since, triangle(PQB)~triangle (ABC)
ar(PQB)/ar(ABC) = PQ2 / BC2 = 1/4
area(PQB) = 1/4 * ar(ABC)
Similarly, ar(SDR) = 1/4*ar(ADC)
ar(CRQ) = 1/4*ar(CDB)
ar(ASP) = 1/4*(ADB)
ar(PQRS) = ar(ABCD) - ar(PQB) - ar(SDR) - ar(CRQ) - ar(ASP)
ar(PQRS) = ar(ABCD) - 1/4* (ar(ABC)+ar(ADC)+ar(CDB)+ar(ADB))
ar(PQRS) = ar(ABCD) - 1/4* (2* ar(ABCD))
ar(PQRS) =1/2 ar(ABCD)