## EXERCISE 8.2

#### Page No 150:

#### Question 1:

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

(i) SR || AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

#### Answer:

(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

∴ SR || AC and SR = AC … (1)

(ii) In ΔABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem,

PQ || AC and PQ = AC … (2)

Using equations (1) and (2), we obtain

PQ || SR and PQ = SR … (3)

⇒ PQ = SR

(iii) From equation (3), we obtained

PQ || SR and PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.

Hence, PQRS is a parallelogram.

#### Question 2:

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.

#### Answer:

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = AC (Using mid-point theorem) … (1)

In ΔADC,

R and S are the mid-points of CD and AD respectively.

∴ RS || AC and RS = AC (Using mid-point theorem) … (2)

From equations (1) and (2), we obtain

PQ || RS and PQ = RS

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to

each other, it is a parallelogram.

Let the diagonals of rhombus ABCD intersect each other at point O.

In quadrilateral OMQN,

MQ || ON ( PQ || AC)

QN || OM ( QR || BD)

Therefore, OMQN is a parallelogram.

⇒ ∠MQN = ∠NOM

⇒ ∠PQR = ∠NOM

However, ∠NOM = 90° (Diagonals of a rhombus are perpendicular to each other)

∴ ∠PQR = 90°

Clearly, PQRS is a parallelogram having one of its interior angles as 90º.

Hence, PQRS is a rectangle.

#### Question 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.

#### Answer:

Let us join AC and BD.

In ΔABC,

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

∴ PQ || AC and PQ = AC (Mid-point theorem) … (1)

Similarly in ΔADC,

SR || AC and SR = AC (Mid-point theorem) … (2)

Clearly, PQ || SR and PQ = SR

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to

each other, it is a parallelogram.

∴ PS || QR and PS = QR (Opposite sides of parallelogram)… (3)

In ΔBCD, Q and R are the mid-points of side BC and CD respectively.

∴ QR || BD and QR =BD (Mid-point theorem) … (4)

However, the diagonals of a rectangle are equal.

∴ AC = BD …(5)

By using equation (1), (2), (3), (4), and (5), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus.

#### Question 4:

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 (see the given figure). Show that F is the mid-point of BC.

#### Answer:

Let EF intersect DB at G.

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.

In ΔABD,

EF || AB and E is the mid-point of AD.

Therefore, G will be the mid-point of DB.

As EF || AB and AB || CD,

∴ EF || CD (Two lines parallel to the same line are parallel to each other)

In ΔBCD, GF || CD and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC.

#### Page No 151:

#### Question 5:

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

#### Answer:

ABCD is a parallelogram.

∴AB || CD

And hence, AE || FC

Again, AB = CD (Opposite sides of parallelogram ABCD)

AB =CD

AE = FC (E and F are mid-points of side AB and CD)

In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.

⇒ AF || EC (Opposite sides of a parallelogram)

In ΔDQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.

⇒ DP = PQ … (1)

Similarly, in ΔAPB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that

Q is the mid-point of PB.

⇒ PQ = QB … (2)

From equations (1) and (2),

DP = PQ = BQ

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

#### Question 6:

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

#### Answer:

Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD.

In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that

SP || BD and SP = BD … (1)

Similarly in ΔBCD,

QR || BD and QR = BD … (2)

From equations (1) and (2), we obtain

SP || QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to

each other. Therefore, SPQR is a parallelogram.

We know that diagonals of a parallelogram bisect each other.

Hence, PR and QS bisect each other.

#### Question 7:

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) MD ⊥ AC

(iii)

#### Answer:

(i) In ΔABC,

It is given that M is the mid-point of AB and MD || BC.

Therefore, D is the mid-point of AC. (Converse of mid-point theorem)

(ii) As DM || CB and AC is a transversal line for them, therefore,

∠MDC + ∠DCB = 180º (Co-interior angles)

∠MDC + 90º = 180º

∠MDC = 90º

∴ MD ⊥ AC

(iii) Join MC.

In ΔAMD and ΔCMD,

AD = CD (D is the mid-point of side AC)

∠ADM = ∠CDM (Each 90º)

DM = DM (Common)

∴ΔAMD ≅ ΔCMD (By SAS congruence rule)

Therefore, AM = CM (By CPCT)

However, AM = AB (M is the mid-point of AB)

Therefore, it can be said that

CM = AM = AB