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)
Solution:
We know that theorem 6.1 states that “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (Basic Proportionality theorem)”.
Consider the diagram as shown below.
In ΔABC, D is the midpoint of AB
Therefore,
AD = BD
AD/BD = 1 .......... (i)
Now,
DE || BC
⇒ AE/EC = AD/BD [Theorem 6.1]
⇒ AE/EC = 1 [From equation (1)]
⇒ AE = EC
Hence, E is the midpoint of AC.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 6
Video Solution:
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)
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 7
Summary:
Hence it is proved that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. Thus, E is the midpoint of AC.
☛ Related Questions:
- 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.
- In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)