NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 3
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Converse of midpoint theorem: It states that in a triangle line drawn from the midpoint of the one side of triangle, parallel to the other side intersect the third side at its at midpoint.
Q5 In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively (see Fig.). Show that the line segments AF and EC trisect the diagonal BD.
Solution:
Given in parallelogram ABCD, E and F are the midpoint of side AB and CD respectively.
Proof:

ABCD is a parallelogram

Therefore,

Also,
Now,

(Opposite sides of parallelogram ABCD)


(E and F are midpoints of side AB and CD)

AECF is a parallelogram (AE and CF are parallel and equal to each other)

(Opposite sides of a parallelogram)
Now, In

F is midpoint of side DC and (as ).

P is the midpoint of DQ (Converse of midpoint theorem)

……..equation (1)
Similarly, In APB,

E is midpoint of side AB and (as).

Q is the midpoint of PB (Converse of midpoint theorem)

…….equation (2)
From equations (1) and (2),


Hence, the line segments AF and EC trisect the diagonal BD.
Q6 Show that the line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other.
Solution:
Given, ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now, In,

R and S are the mid points of CD and DA respectively.

.
Similarly we can show that,
Thus, PQRS is parallelogram and PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.