# O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO).

**Solution:**

Given, PQRS is a parallelogram

PR is the diagonal

O is any point on the diagonal PR

We have to prove that ar(PSO) = ar(PQO)

Join SQ which intersects PR at B.

We know that the diagonals of the parallelogram bisect each other.

So, B is the midpoint of SQ

We know that the median of a triangle divides it into two triangles of equal areas.

PB is a median of triangle QPS

ar(BPQ) = ar(BPS) -------------------- (1)

OB is a median of triangle OSQ

ar(OBQ) = ar(OBS) ---------------------- (2)

Adding (1) and (2),

ar(BPQ) + ar(OBQ) = ar(BPS) + ar(OBS)

From the figure,

BPQ + BOQ = PQO

ar(POQ) = ar(BPS) + ar(OBS)

From the figure,

BPS + OBS = PSO

Therefore, ar(PQO) = ar(PSO)

**✦ Try This: **D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC. Show that ar(BDEF) = 1/2 ar(ABC)

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 6**

## O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO)

**Summary:**

O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). It is proven that ar (PSO) = ar (PQO)

**☛ Related Questions:**

- ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD . . . .
- In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn . . . .
- If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the pa . . . .

visual curriculum