*Q1.*

(i)

(ii) PQ = SR

(iii) PQRS is a parallelogram.

**Solution:**

(i) In

R is the mid point of DC and S is the mid point of DA.

Thus by mid point theorem,

(ii) In

P is the mid point of AB and Q is the mid point of BC.

Thus by mid point theorem,

also,

Thus, PQ = SR

(iii)

and,

also, PQ = SR

Thus, PQRS is a parallelogram.

*Q2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.*

**Solution:**

**Given,**

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

**To Prove,**

PQRS is a rectangle.

**Construction,**

Join AC and BD

**Proof,**

In

DS = BQ (Halves of the opposite sides of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

Thus,

RS = PQ (by CPCT) ………… (i)

In

RC = PA (Halves of the opposite sides of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

Thus,

RQ = SP (by CPCT) ……(ii)

Now,

In

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

also,

P and S are the mid points of AD and AB respectively.

Thus, PQRS is a parallelogram.

also,

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

Thus, PQRS is a rectangle.

*Q3. 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 a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

**Construction,**

AC and BD are joined.

**To Prove,**

PQRS is a rhombus.

**Proof,**

In

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

Thus,

In

So,

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.

Now,

In

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

Thus,

AC = BD (Diagonals of a rectangle are equal) ……… (v)

From equations (i), (ii), (iii), (iv) and (v),

PQ = QR = SR = PS

So, PQRS is a rhombus.

*Q4. ABCD is a trapezium in which AB∥DC, 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. Show that F is the mid-point of BC.*

** **

**Solution:**

**Given,**

ABCD is a trapezium in which

**To prove,**

F is the mid-point of BC.

**Proof,**

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 mid point theorem)

Now,

In

G is the mid point of BD and also

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

*Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.*

* *

**Solution:**

**Given**

ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.

**To show,**

AF and EC trisect the diagonal BD.

**Proof,**

ABCD is a parallelogram

Therefore,

also,

Now,

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

Now,

In

F is mid point of side DC and

P is the mid-point of DQ (Converse of mid-point theorem)

Similarly,

In

E is mid point of side AB and

Q is the mid-point of PB (Converse of mid-point theorem)

From equations (i) and (i),

DP = PQ = BQ

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

*Q6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
*

**Solution:**

Let 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.

Thus,

Similarly we can show that,

Thus, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

*Q7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) *

**Solution:**

(i) In

M is the mid point of AB and

Thus, D is the mid point of AC (Converse of mid point theorem)

(ii)

also,

Thus,

(iii) In

AD = CD (D is the midpoint of side AC)

DM = DM (common)

Thus,

AM = CM (by CPCT)

also,

Hence,