In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS. Show that :

(i) 2 Area (△ POS) = Area (// gm PMLS)

(ii) Area (△ POS) + Area (△ QOR) = $\dfrac{1}{2}$ Area (//gm PQRS)

(iii) Area (△ POS) + Area (△ QOR) = Area (△ POQ) + Area (△ SOR)

(i) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

Triangle POS and parallelogram PMLS lie on same base PS and between same parallel lines PS and ML.

∴ Area of triangle POS = $\dfrac{1}{2}$ Area of parallelogram PMLS

⇒ Area of parallelogram PMLS = 2 (Area of triangle POS)

Hence, proved that 2 area (△ POS) = area (// gm PMLS).

(ii) We know that,

Area of a triangle is half that of a parallelogram on the same base and between the same parallels.

From figure,

Triangle POS and parallelogram PMLS lie on same base PS and between same parallel lines PS and ML.

∴ Area of triangle POS = $\dfrac{1}{2}$ Area of parallelogram PMLS ……….(1)

Triangle QOR and parallelogram MQRL lie on same base QR and between same parallel lines QR and ML.

∴ Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram MQRL ……….(2)

Adding equations (1) and (2), we get :

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram PMLS + $\dfrac{1}{2}$ Area of parallelogram MQRL

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ (Area of parallelogram PMLS + Area of parallelogram MQRL)

⇒ Area of triangle POS + Area of triangle QOR = $\dfrac{1}{2}$ Area of parallelogram PQRS.

Hence, proved that area (△ POS) + area (△ QOR) = $\dfrac{1}{2}$ area (//gm PQRS).

(iii) We know that,

In a parallelogram diagonals bisect each other.

∴ OS = OQ

In △ PQS,

O is the mid-point of QS.

∴ OP is the median.

We know that,

Median of a triangle divides it into two triangles of equal areas.

∴ Area of △ POS = Area of △ POQ ………(3)

In △ QSR,

O is the mid-point of QS.

∴ OR is the median.

We know that,

Median of a triangle divides it into two triangles of equal areas.

∴ Area of △ QOR = Area of △ SOR ………(4)

Adding equations (3) and (4), we get :

⇒ Area of △ POS + Area of △ QOR = Area of △ POQ + Area of △ SOR.

Hence, proved that area (△ POS) + area (△ QOR) = area (△ POQ) + area (△ SOR).