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**Exercise 8.2****1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that : (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram.**

**Answer**

(i) In Î”DAC,

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

Thus by mid point theorem, SR || AC and SR = 1/2 AC

(ii) In Î”BAC,

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

Thus by mid point theorem, PQ || AC and PQ = 1/2 AC

also, SR = 1/2 AC

Thus, PQ = SR

(iii) SR || AC - from (i)

and, PQ || AC - from (ii)

â‡’ SR || PQ - from (i) and (ii)

also, PQ = SR

Thus, PQRS is a parallelogram.**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 **

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,

AC and BD are joined.

Proof,

In Î”DRS and Î”BPQ,

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

âˆ SDR = âˆ QBP (Opposite angles of the rhombus)

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

Thus, Î”DRS â‰… Î”BPQ by SAS congruence condition.

RS = PQ by CPCT --- (i)

In Î”QCR and Î”SAP,

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

âˆ RCQ = âˆ PAS (Opposite angles of the rhombus)

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

Thus, Î”QCR â‰… Î”SAP by SAS congruence condition.

RQ = SP by CPCT --- (ii)

Now,

In Î”CDB,

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

â‡’ QR || BD

also,

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

â‡’ PS || BD

â‡’ QR || PS

Thus, PQRS is a parallelogram.

also, âˆ PQR = 90Â°

Now,

In PQRS,

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

âˆ Q = 90Â°

Thus, PQRS is a rectangle.**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 **

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 Î”ABC

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

Thus, PQ || AC and PQ = 1/2 AC (Mid point theorem) --- (i)

In Î”ADC,

SR || AC and SR = 1/2 AC (Mid point theorem) --- (ii)

So, PQ || SR and PQ = SR

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

PS || QR and PS = QR (Opposite sides of parallelogram) --- (iii)

Now,

In Î”BCD,

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

Thus, QR || BD and QR = 1/2 BD (Mid point theorem) --- (iv)

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.**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 Fig. 8.30). Show that F is the mid-point of BC.**

**Answer**

Given,

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.

To prove,

F is the mid-point of BC.

Proof,

BD intersected EF at G.

In Î”BAD,

E is the mid point of AD and also EG || AB.

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

Now,

In Î”BDC,

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

Thus, F is the mid point of BC (Converse of mid point theorem)**5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.**

**Answer**

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

Therefor, AB || CD

also, AE || FC

Now,

AB = CD (Opposite sides of parallelogram ABCD)

â‡’ 1/2 AB = 1/2 CD

â‡’ AE = FC (E and F are midpoints of side AB and CD)

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

AF || EC (Opposite sides of a parallelogram)

Now,

In Î”DQC,

F is mid point of side DC and FP || CQ (as AF || EC).

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

â‡’ DP = PQ --- (i)

Similarly,

In APB,

E is mid point of side AB and EQ || AP (as AF || EC).

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

â‡’ PQ = QB --- (ii)

From equations (i) and (i),

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.**6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Answer **

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Now,

In Î”ACD,

R and S are the mid points of CD and DA respectively.

Thus, SR || AC.

Similarly we can show that,

PQ || AC

PS || BD

QR || BD

Thus, PQRS is parallelogram.

PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.**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 âŠ¥ AC (iii) CM = MA = 1/2 AB**

(i) In Î”ACB,

M is the mid point of AB and MD || BC

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

(ii) âˆ ACB = âˆ ADM (Corresponding angles)

also, âˆ ACB = 90Â°

Thus, âˆ ADM = 90Â° and MD âŠ¥ AC

(iii) In Î”AMD and Î”CMD,

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

âˆ ADM = âˆ CDM (Each 90Â°)

DM = DM (common)

Thus, Î”AMD â‰… Î”CMD by SAS congruence condition.

AM = CM by CPCT

also, AM = 1/2 AB (M is mid point of AB)

Hence, CM = MA = 1/2 AB

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