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# In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

**Solution:**

We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

In ΔPOQ

DE || OQ (given)

PE/EQ = PD/DO..........(1)

In ΔPOR

DF || OR ( given)

PF/FR = PD/DO......... (2)

From equation (1) and (2)

PE/EQ = PF/FR = PD/DO

PE/EQ = PF/FR

In ΔPQR

PE/EQ = PF/FR

∴ QR || EF (Converse of Basic Proportionality theorem)

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 5

**Summary:**

Hence it is proved EF || QR if in the above figure when DE || OQ and DF || OR.

**☛ Related Questions:**

- In Fig. 6.21, A, B and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
- Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
- Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
- ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

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