NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 2
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Midpoint
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it
Q3 ABCD is a rectangle and P, Q, R and S are midpoints of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:

Give, ABCD is a ABCD is a rectangle and PQRS is a rhombus quadrilateral .P, Q, R

And S is midpoints of the sides AB, BC, CD and DA respectively.
In

P and Q are the midpoints of AB and BC respectively

Thus, and (Midpoint theorem) …….equation (1)
In,

S, R are the midpoints of AD and DC respectively

and (Midpoint theorem)……….equation (2)
From equation (1) and (2)
So and
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
and (Opposite sides of parallelogram) ……..equation (3)
Now, In,

Q and R are mid points of side BC and CD respectively.

Thus, and (Midpoint theorem) ……..equation (4)

(Diagonals of a rectangle are equal)………..equation (5)
From equations (1), (2), (3), (4) and (5),


So, PQRS is a rhombus.
Q4 ABCD is a trapezium in which, BD is a diagonal and E is the midpoint of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig.). Show that F is the midpoint of BC.
Solution:

Give, ABCD is a trapezium in which,

BD is a diagonal and E is the midpoint of AD.

A line is drawn through E parallel to AB intersecting BC at F
Proof

F is the midpoint of BC.

BD intersected EF at G.

In ,


E is the midpoint of AD and also

Thus, G is the midpoint of BD (Converse of midpoint theorem)
Now, In ΔBDC,

G is the midpoint of BD and also GF  AB  DC.

Thus, F is the midpoint of BC (Converse of midpoint theorem)