# NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 3 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Converse of mid-point theorem: It states that in a triangle line drawn from the mid-point of the one side of triangle, parallel to the other side intersect the third side at its at mid-point.

Q-5 In a parallelogram ABCD, E and F are the mid-points 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 mid-point 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 mid-point of DQ (Converse of mid-point theorem)
- … equation (1)

Similarly, In APB,

- E is midpoint of side AB and (as ) .
- Q is the mid-point of PB (Converse of mid-point theorem)
- … equation (2)

From equations (1) and (2) ,

- Hence, the line segments AF and EC trisect the diagonal BD.

Q-6 Show that the line segments joining the mid-points 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.