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:
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 ΔOPQ,
AB || PQ (given)
OA/AP = OB/BQ .............. (i) [By Basic proportionality theorem]
In ΔOPR
AC || PQ (given)
OA/AP = OC/CR............. (ii) [By Basic proportionality theorem]
From equations (i) and (ii)
OA/AP = OB/BQ = OC/CR
OB/BQ = OC/CR
Now, In ΔOQR
OB/BQ = OC/CR
Thus, BC || QR [By Converse of Basic proportionality theorem]
☛ Check: NCERT Solutions Class 10 Maths Chapter 6
Video Solution:
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.
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 6
Summary:
Hence it is proved BC || QR if A, B and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR.
☛ Related 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).
- 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.
- The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.