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Question 6 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.

Solution:

Given,

In ΔOPQ, AB || PQ

By using Basic Proportionality Theorem,

$\dfrac{OA}{AP} = \dfrac{OB}{BQ}$…………….(i)

Also given,

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

∴ $\dfrac{OA}{AP} = \dfrac{OC}{CR}$……………(ii)

From equation (i) and (ii), we get,

$\dfrac{OB}{BQ} = \dfrac{OC}{CR}$

Therefore, by converse of Basic Proportionality Theorem,

In ΔOQR, BC || QR.

Additional Questions

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).

Answer

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).

Answer

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$

Answer

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$ Show that ABCD is a trapezium.

Answer

In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Answer

E and F are points on the sides PQ and PR respectively of a Δ PQR.
For each of the following cases, state whether EF || QR :

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Answer

In Fig. 6.18, if LM || CB and LN || CD, prove that $\dfrac{AM}{AB} = \dfrac{ AN}{AD}$

Answer

In Fig. 6.19, DE || AC and DF || AE. Prove that $\dfrac{BF}{FE} = {BE}{EC}$

Answer

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

Answer

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.

Answer
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