Using 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.
=> $$PQ = frac{1}{2}AC$$ and PQ | | AC Similarly, $$RS = frac{1}{2}AC$$ and RS | | AC => PQRS is a parallelogram. Also, $$ riangle$$PQB $$sim riangle$$ABC => $$frac{ar ( riangle PQB)}{ar ( riangle ABC)} = frac{PQ^2}{BC^2} = frac{1}{4}$$ => $$ar( riangle PQB) = frac{1}{4} ar( riangle ABC)$$ SImilarly, $$ar ( riangle SDR) = frac{1}{4} ar ( riangle ADC)$$ $$ar ( riangle CRQ) = frac{1}{4} ar ( riangle CDB)$$ $$ar ( riangle ASP) = frac{1}{4} ar ( riangle ADB)$$ => $$ar(PQRS) = ar(ABCD) - ar ( riangle PQB) - ar( riangle SDR) - ar ( riangle CRQ) - ar( riangle ASP)$$ => $$ar(PQRS) = ar(ABCD) - frac{1}{4} * [ar( riangle ABC) + ar( riangle ADC) + ar( riangle CDB) + ar( riangle ADB)]$$ => $$ar(PQRS) = ar(ABCD) - frac{1}{4} * [2 * ar(ABCD)]$$ => $$ar(PQRS) = frac{1}{2} ar(ABCD)$$