NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 3
Converse of mid-point theorem: 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.
ABCD is a parallelogram
(Opposite sides of parallelogram ABCD)
(E and F are midpoints of side AB and CD)
Therefore, AECF is a parallelogram (AE and CF are parallel and equal to each other)
(Opposite sides of a parallelogram)
Similarly, In APB,
E is midpoint of side AB and (as ).
Therefore, Q is the mid-point of PB (Converse of mid-point theorem)
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.
ABCD is a quadrilateral
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.
Now, in ,
Similarly we can show that,
Thus, PQRS is parallelogram.
Since, PR and QS are the diagonals of the parallelogram PQRS. So, they bisect each other.