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)
Solution:
We know that theorem 6.2 tells us if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of Basic Proportionality theorem)
In ΔABC,
D is the midpoint of AB
⇒ AD = BD
AD/BD = 1............ (i)
E is the midpoint of AC
AE = CE
⇒ AE/CE = 1........ (ii)
From equations (i) and (ii)
AD/BD = AE/CE = 1
AD/BD = AE/CE
In ΔABC, according to theorem 6.2 (Converse of Basic Proportionality theorem),
Since, AD/BD = AE/CE
Thus, DE || BC
Hence, proved.
☛ Check: NCERT Solutions Class 10 Maths Chapter 6
Video Solution:
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)
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 8
Summary:
Using Theorem 6.2, we have proved that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
☛ Related Questions:
- 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.
- In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
- 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.