# D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is

a. a square

b. a rectangle

c. a rhombus

d. a parallelogram

**Solution:**

In Δ ABC

D and E are the mid-points of AB and AC

From the mid-point theorem

DE || BC

DE = 1/2 BC

So we get

DE = 1/2 [BP + PO + OQ + QC]

DE = 1/2 [2PO + 2OQ]

As P and Q are the midpoints of OB and OC

DE = PO + OQ

DE = PQ

In Δ AOC

Q and C are the midpoints of AC and OC

In Δ AOB

PD || AO

PD = 1/2 AO [Using mid-point theorem]

From Δ AOC and Δ AOB

EQ || PD and EQ = PD

From Δ ABC

DE || BC or DE || PQ

DE = PQ

So DEQP is a parallelogram

Therefore, if P and Q are the mid-points of OB and OC respectively, then DEQP is a parallelogram.

**✦ Try This: **The figure obtained by joining the mid-points of the adjacent sides 10 cm and 5 cm a. a rhombus, b. a rectangle, c. a square, d. any parallelogram

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 8

**NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 10**

## D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is , a. a square, b. a rectangle, c. a rhombus, d. a parallelogram

**Summary:**

D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is a parallelogram

**☛ Related Questions:**

- The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is . . . .
- The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32º a . . . .
- Which of the following is not true for a parallelogram? ,a. opposite sides are equal ,b. opposite an . . . .

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