## 8. QUADRILATERALS

8.1 Introduction

8.2 Angle Sum Property of a Quadrilateral

8.3 Types of Quadrilaterals

8.4 Properties of a Parallelogram

8.5 Another Condition for a Quadrilteral to be a Parallelogram

8.6 The Mid-point Theorem

8.7 Summary

**Important formula**

*Exercise 8.1*

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Q.1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Let x be the common ratio between the angles.

Sum of the interior angles of the quadrilateral = 360°

Now,

3x + 5x + 9x + 13x = 360°

⇒ 30x = 360°

⇒ x = 12°

Angles of the quadrilateral are:

3x = 3×12° = 36°

5x = 5×12° = 60°

9x = 9×12° = 108°

13x = 13×12° = 156°

Q.2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Given,

AC = BD

To show,

To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.

Proof,

In ΔABC and ΔBAD,

BC = BA (Common)

AC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.

∠A = ∠B (by CPCT)

also,

∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠A = 180°

⇒ ∠A = 90° = ∠B

Thus ABCD is a rectangle.

Q.3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given,

OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

To show,

ABCD is parallelogram and AB = BC = CD = AD

Proof,

In ΔAOB and ΔCOB,

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.

Thus, AB = BC (by CPCT)

Similarly we can prove,

AB = BC = CD = AD

Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Q.4. Show that the diagonals of a square are equal and bisect each other at right angles.

Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To show,

AC = BD, AO = OC and ∠AOB = 90°

Proof,

In ΔABC and ΔBAD,

BC = BA (Common)

∠ABC = ∠BAD = 90°

AC = AD (Given)

Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.

Thus, AC = BD by CPCT. Therefore, diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.

Thus, AO = CO by CPCT. (Diagonal bisect each other.)

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.

also, ∠AOB = ∠COB

∠AOB + ∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)

Q.5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Given,

Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angle at O.

To prove,

Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.

Thus, AB = CD by CPCT. — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Vertically opposite)

OD = OD (Common)

Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.

Thus, AD = CD by CPCT. — (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB — (ii)

also, ∠ADC = ∠BCD by CPCT.

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° — (iii)

One of the interior ang is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

Q.6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

(i)

In ΔADC and ΔCBA,

AD = CB (Opposite sides of a ||gm)

DC = BA (Opposite sides of a ||gm)

AC = CA (Common)

Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.

Thus,

∠ACD = ∠CAB by CPCT

and ∠CAB = ∠CAD (Given)

⇒ ∠ACD = ∠BCA

Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)

⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)

Also, AB = BC = CD = DA (Opposite sides of a ||gm)

Thus, ABCD is a rhombus.

Q.7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Let ABCD is a rhombus and AC and BD are its diagonals.

Proof,

AD = CD (Sides of a rhombus)

∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)

also, AB || CD

⇒ ∠DAC = ∠BCA (Alternate interior angles)

⇒ ∠DCA = ∠BCA

Therefore, AC bisects ∠C.

Similarly, we can prove that diagonal AC bisects ∠A.

Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.

Q.8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:

(i) ABCD is a square

(ii) diagonal BD bisects ∠B as well as ∠D.

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

Therefore, AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In ΔBCD,

BC = CD

⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)

also, ∠CDB = ∠ABD (Alternate interior angles)

⇒ ∠CBD = ∠ABD

Thus, BD bisects ∠B

Now,

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

Thus, BD bisects ∠D

Q.9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

(i) In ΔAPD and ΔCQB,

DP = BQ (Given)

∠ADP = ∠CBQ (Alternate interior angles)

AD = BC (Opposite sides of a ||gm)

Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,

BQ = DP (Given)

∠ABQ = ∠CDP (Alternate interior angles)

AB = BCCD (Opposite sides of a ||gm)

Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

Q.10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

(i) In ΔAPB and ΔCQD,

∠ABP = ∠CDQ (Alternate interior angles)

∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

Q.11. In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

(i) AB = DE and AB || DE (Given)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

⇒ AD = BE and BE = CF (Opposite sides of a parallelogram are equal)

Thus, AD = CF.

Also, AD || BE and BE || CF (Opposite sides of a parallelogram are parallel)

Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ΔABC ≅ ΔDEF by SSS congruence condition.

Q.

12. ABCD is a trapezium in which AB || CD and

AD = BC (see Fig. 8.23). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(iv) diagonal AC = diagonal BD

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

⇒ ∠CBE = ∠CEB

also,

∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)

∠B + ∠CBE = 180° (Linear pair)

⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)

⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)

⇒ ∠D = ∠C

(iii) In ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA

AD = BC (Given)

Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

*Exercise 8.2*

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Q.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.

(i) In ΔDAC,

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 ΔBAC,

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)

⇒ SR || PQ – from (i) and (ii)

also, PQ = SR

Thus, PQRS is a parallelogram.

Q.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.

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 ΔDRS and ΔBPQ,

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

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

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

Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.

RS = PQ by CPCT — (i)

In ΔQCR and ΔSAP,

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

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

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

Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.

RQ = SP by CPCT — (ii)

Now,

In ΔCDB,

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

⇒ QR || BD

also,

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

⇒ PS || BD

⇒ QR || PS

Thus, PQRS is a parallelogram.

also, ∠PQR = 90°

Now,

In PQRS,

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

∠Q = 90°

Thus, PQRS is a rectangle.

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.

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 ΔABC

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 ΔADC,

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 ΔBCD,

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.

Q.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.

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 ΔBAD,

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 ΔBDC,

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)

Q.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.

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)

⇒ 1/2 AB = 1/2 CD

⇒ 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 ΔDQC,

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)

⇒ 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)

⇒ PQ = QB — (ii)

From equations (i) and (i),

DP = PQ = BQ

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

Q.6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

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

Now,

In ΔACD,

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.

Q.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) CM = MA = 1/2 AB

(i) In ΔACB,

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

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

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

also, ∠ACB = 90°

Thus, ∠ADM = 90° and MD ⊥ AC

(iii) In ΔAMD and ΔCMD,

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

∠ADM = ∠CDM (Each 90°)

DM = DM (common)

Thus, ΔAMD ≅ ΔCMD 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

Maths NCERT solutions class 9 Chapter 8 QUADRILATERALS

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