Mid-point Theorem: 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.

Q-3 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Solution:

Given,

ABCD is ABCD is a rectangle and PQRS is a quadrilateral formed by joining midpoints of its sides.

In

P and Q are the mid-points of AB and BC respectively

Thus, and (Mid-point theorem) …….equation (1)

In,

S, R are the mid-points of AD and DC respectively

Therefore, and (Mid-point theorem)……….equation (2)

From equation (1) and (2)

and

Therefore, in quadrilateral PQRS pair of opposite sides is equal and parallel, so, it is a parallelogram.

Therefore, and (Opposite sides of parallelogram) ……..equation (3)

Now, In,

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

Thus, and (Mid-point theorem) ……..equation (4)

(Diagonals of a rectangle are equal)………..equation (5)

From equations (1), (2), (3), (4) and (5),

Since PQRS is a parallelogram, with equal sides. Therefore, PQRS is a rhombus.

Q-4 ABCD is a trapezium in which, , BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersects BC at F (see Fig.). Show that F is the mid-point of BC.

Solution:

Given,

ABCD is a trapezium in which ,

BD is a diagonal and E is the mid-point of AD.

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

To prove: F is the mid-point of BC.

Proof,

Now, BD intersected EF at G.

In ,

E is the mid-point of AD and also

Thus, G is the mid-point of BD (Converse of midpoint theorem)

Now, in ΔBDC,

G is the mid-point of BD and also GF AB DC.

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