Chapter 9 Quadrilaterals

**NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.1**

**Question 1.**

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

**Solution:**

**Given:** the ratio of the angles of quadrilateral are 3 : 5 : 9 : 13.

Let the angles of the quadrilateral are 3x, 5x, 9x and 13x.

We know that, sum of angles of a quadrilateral = 360°

∴ 3x + 5x+ 9x+13x = 360°

⇒ 30x = 360° ⇒ x = 360∘30 = 12°

∴ Angles of the quadrilateral are 3x = 3 x 12 = 36°

5x = 5 x 12 = 60°

9x= 9×12 = 108°

and 13x = 13×12 = 156°

**Question 2.**

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

**Solution:**

Let given parallelogram is ABCD whose diagonals AC and BD are equal, i.e., AC = BD.

Now, we have to prove that ABCD is a rectangle.

**Proof:** In ∆ ABC and ∆ DCB, we have

AB = CD (Opposite sides of parallelogram)

BC = CB (Common in both triangles)

and AC = BD (Given)

∴ ∆ABC ≅ ∆DCB (By SSS rule)

∴ ∆ABC = ∠DCB …(i)

(Corresponding Part of Congruent Triangle)

But DC || AB and transversal CB intersect them.

∴ ∠ABC+ ∠DCB = 180°

(∵ Both are interior angles on the same side of the transversal)

⇒ ∠ABC + ∠ABC = 180° [From Eq. (i)]

⇒ 2 ∠ ABC = 180°

⇒ ∠ABC = 90°= ∠DCB

Thus, ABCD is a parallelogram and one of angles is 90°.

Hence, ABCD is a rectangle.

Hence proved.

**Question 3.**

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

**Solution:**

**Given:** a quadrilateral ABCD whose diagonals AC and BD bisect each other at right angles.

i.e, OA = OC and OB = OD

and ∠AOD = ∠AOB = ∠COD = ∠BOC = 90°

**To prove:** ABCD is a rhombus.

**Proof:** In ∆OAB and ∆ODC, we have

OA = OC and OB = OD (Given)

∠AOB = ∠COD (Vertically opposite angles)

∴ ∆OAB ≅ ∆OCD (By SAS rule)

∴ AB = CD ,..(i)

(Corresponding part of congruent triangles)

Again, in ∆OAD and ∆OBC, we have

OA = OC and OD = OB (Given)

and ∠AOD = ∠BOC (Vertically opposite angle)

∴ A OAD = A OCB (By SAS rule)

∴ AD = BC …(ii) (Corresponding part of congruent triangles)

Similarly, we can prove that

AB = AD

CD = BC …(iii)

Hence, from Eqs. (i), (ii) and (iii), we get

AB = BC = AD = CD

Hence, ABCD is a rhombus.

Hence proved.

**Question 4.**

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

**Solution:**

**Given:** A square ABCD whose diagonals AC and BD intersect at O.

i. e, . AC = BD, OD = OB, OA = OC and AC ⊥ BD

**Proof:** In A ABC and A BAD, we have

AB = BA (Common)

BC= AD (Sides of a square)

∠ABC = ∠BAD = 90°

∴ ∆ABC ≅ ∆BAD (By SAS rule)

AC = BD

(Corresponding Parts of Congruent Triangle)

∴ ∆OAB and ∆OCD

AB = DC (Sides of a square)

∠ OAB = ∠DCO

(∵ AB || CD and transversal AC intersect)

and ∠ OBA = ∠BDC

(∵AB|| CD and transversal BD intersect)

∴ ∆OAB ≅ ∆OCD

OA = OC and OB = OD

(Corresponding Parts of Congruent Triangle)

Now, in ∆AOB and ∆AOD, we have

OB = OD (Prove in above)

AB = AD (Sides of a square)

AO = OA (Common)

∴ ∆ AOB = ∆AOD (By SSS)

∴ ∠AOB = ∠AOD (By CPCT)

But ∠AOB + ∠AOD-180° (Linear pair)

∴ ∠AOB = ∠AOD = 90°

Thus, AO ⊥ BDi.e.,AC ⊥ BD.

Hence, AC = BD, OA = OC, OB = OD and AC ⊥ BD

Hence proved.

**Question 5.**

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

**Solution:**

**Given:** A quadrilateral ABCD in which AC = BD and AC ⊥ BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.

**To prove:** ABCD is a square.

**Proof:** Let AC and BD intersect at a point O.

In ∆ABO and ∆ADO, we have

BO = OD (Given)

AO = OA (Common)

∠AOB = ∠AOD = 90° (Given)

∴ ∆ABO ≅ ∆ADO (By SAS)

∴ AB = AD (By CPCT)

and AD = BC (Opposite sides of parallelogram)

∴ AB = BC = DC = AD …(i)

Again, in ∆ABC and ∆BAD, we have

AB = BA (Common)

AC = BD (Given)

BC= AD [From Eq. (i)]

∴ ∆ABC = ∆BAD (By SSS)

∠ABC= ∠ BAD

∠ABC+ ∠ BAD =180° …(ii)

But

∠ABC = ∠ BAD = 180° (Sum of interior angles of a parallelogram)

∴ ∠ABC = ∠BAD = 90° [From Eq. (ii)]

Thus, AB = BC = CD = DA and ∠A = 90°

∴ ABCD is a square.

Hence proved.

**Question 6.**

**Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that**

**(i) it bisects ∠C also,**

**(ii) ABCD is a rhombus.**

**Solution:**

**Given:** diagonal AC of a parallelogram ABCD bisects ∠A

i.e., ∠DCA = ∠ BAC = 12 ∠BAD ….(i)

Here, AB || CD and AC is transversal.

∴ ∠DCA = ∠CAB (Pair of alternate angle)…(ii)

and ∠BCA = ∠ DAC (Pair of alternate angle).. .(iii)

From Eqs. (i), (ii) and (iii), we get

∠DAC = BCA = ∠ BAC = ∠DCA

Now, ∠BCD = ∠BCA + ∠DCA

= ∠DAC + ∠CAB

= ∠BAD

∴ Diagonal AC also bisects ∠C.

Again, in ∆OAD and ∆ OCD, we have

OA = OC (∵ Diagonals bisect each other)

OD = OD (Common)

∴ ∠AOD = ∠ COD = 90°

∴ ∆ OAD ≅ ∆ OCD (By SAS)

∴ AD = CD (By CPCT)

Now AB = CD and AD = BC (Opposite sides of parallelogram)

∴ AB = CD = AD = BC

Hence, ABCD is a rhombus.

Hence proved.

**Question 7.**

**ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.**

**Solution:**

**Given:** ABCD is a rhombus

∴ AD = AB = BC = CD ….(i)

**To prove:**

(i) Diagonal AC bisect ∠A as well as ∠C.

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

**Proof:**

(i) Let AC and BD are the diagonals of rhombus ABCD.

In ∆ABC and ∆ADC,

AD = AB

CD = BC [From Eq,(i)]

and AC = CA (Common)

∴ ∆ABC ≅ ∆ADC ( By SSS rule)

∴ ∠DAC = ∠BAC (By CPCT)

and ∠DCA = ∠ BCA

Also, ∠DAC = ∠DCA

and ∠BAC = ∠ BCA

This shows that AC bisect ∠ A as well as ∠C.

(ii) Again, in ∆BDCand ∆BDA,

AB = BC

AD = CD

BD = BD (Common)

∴ ∆ BDC ≅ ∆ BDA (SSS rule)

∴ ∠BDA = ∠BDC

and ∠DBA = ∠DBC

Also, ∠BDA = ∠DBA

and ∠BDC = ∠DBC

This shows that BD bisect ∠B as well as ∠D.

**Question 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.**

**Solution:**

**Given:** ABCD is a rectangle.

∴ AB = CD and BC= AD …(i)

To Prove (i) ABCD is a square.

i.e., AB = BC = CD = DA

(ii) Dioagonal BDbisects ∠B as well as ∠D.

Proof (i) In ∆ADC and ∆ABC, we have

Since, AB || DC and AC transversal intersect

∠DAC = ∠BAC

∠DCA = ∠BCA

and AC = CA (Common)

∴ ∆ADC ≅ ∆ ABC (By ASA rule)

AD = AB (By CPCT)

and CD = BC …(ii)

Hence, from Eqs. (i) and (ii), we get

AB = BC- AD = CD

∴ ABCD is a square.

(ii) In ∆AOB and ∆COB, we have

AB = BC (Side od square)

BO = OB (Common)

OA = OC

(∵ Diagonal of square bisect each other)

∴ ∆AOB ≅ ∆COB

∴ ∠OBA = ∠OBC

This shows that BO or BD bisect ∠B.

Similarly, in ∆ AOD and ∆ COD, we have

AD = CD (Side od square)

OD = DO (Common)

and OA = OC

(∵ Diagonal of square bisect each other)

∴ ∆ AOD ≅ ∆ COD (By ASA rule)

∴ ∠ADO = ∠CDO

This shows that DO or DB bisect ∠D.

Hence proved.

**Question 9.**

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

**Solution:**

**Given:** ABCD is a parallelogram and P and Q are lie on BD such that

DP = PQ ….(i)

(i) We have to show,

∆APD ≅ ∆CQB

Now, in ∆APD and ∆CQB, we have

DP = BQ (Given)

AD = BC

(Opposite sides are equal in parallelogram)

∵ AD || BC and BD is a transversal.

∴ ∠ADP = ∠QBC (Alternate interior angle)

∴ ∆APD ≅ ∆CQB (By SAS)

(ii) Since, ∆APD ≅ ∆CQB

∴ AP = CQ

(iii) Here, we have to show, ∆AQB ≅ ∆CPD

Now, in ∆AQB and ∆CPD, we have

BQ = DP (Given)

AB = CD (Opposite sides of parallelogram)

∵ AB || CD and BD is a transversal.

∴ ∠ABQ = ∠CDP (Alternate interior angle)

∴ ∆AQB s ∆ CPD

(iv) Since, ∆AQB ≅ ∆CPD

∴ AQ = CP

(v) Now, in ∆APQ and ∆PCQ, we have

AQ = CP [From part (iv)]

AP = CQ [From part (ii)]

PQ = QP (Common)

∴ ∆APQ = ∆PCQ (By SSS)

∴ ∠APQ = ∠PQC

and ∠AQP = ∠CPQ (Vertically opposite)

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.

∴ AP || CQ and AQ || CP

Now, both pairs of opposite sides of quadrilateral APCQ are parallel.

Hence, APCQ is a parallelogram.

Hence proved.

**Question 10.**

**ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that**

**Solution:**

**Given:** ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

∵ AB || CD and BD is a transversal.

∴ ∠CDB = ∠DBA …(i)

(i) Now, in ∆ APB and ∆ CQD, we have

CD = AB (Sides of parallelogram)

∠CQD = ∠APB = 90° (Given)

∠CDQ = ∠ABP [From Eq. (i)]

∴ ∆APB ≅ ∆CQD (By ASA rule)

(ii) ∵ ∆APB ≅ ∆CQU (By CPCT)

∴ AP = CQ

Hence proved.

**Question 11.**

**In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).**

**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) ∆ABC ≅ ∆DEF**

**Solution:**

**Given:** in ∆ABC and ∆DEF,

AB = DE, AB || DE

and BC = EF,BC||EF

**
(i)** Now, in quadrilateral ABED,

AB = DE and AB || DE (Given)

⇒ ABED is a parallelogram.

(∵ A pair of opposite sides is equal and parallel)

**In quadrilateral BEFC,**

(ii)

(ii)

BC = EF and BC || EF.

⇒ BEFC is a parallelogram.

(∵ A pair of opposite sides is equal and parallel)

**Since, ABED is a parallelogram.**

(iii)

(iii)

∴ AD || BE and AD = BE …(i)

Also, BEFC in a parallelogram.

∴ CF || BE and CF = BE …(ii)

From Eqs. (i) and (ii), we get

AD || CF and AD =CF

**In quadrilateral ACFD, we have**

(iv)

(iv)

AD || CF and AD = CF [From part (iii)]

⇒ ACFD is a parallelogram.

**Since, ACFD is a parallelogram.**

(v)

(v)

∴ AC = DF and AC || DF

**Now, in ∆ABC and ∆DEF,**

(vi)

(vi)

AB = DE (Given)

BC = EE (Given)

and AC = DF [From part (v)]

∴ ∆ABC ≅ ∆DEF (By SSS rule)

**Question 12.**

**ABCD is a trape∠ium in which AB || CD and AD = BC (see figure). Show that**

**(i )∠A=∠B**

**(ii )∠C=∠D**

**(iii) ∆ABC ≅ ∆BAD**

**(iv) diagonal AC = diagonal BD**

**[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].**

**Solution:**

**Given:** ABCD is a trapezium.

AB || CD and AD = BC

Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Now, ADCE is a parallelogram.

∴ AD || CE and AD = CE

But AD = BC

∴ AD = BC = CE

**
(i)** We know that, ∠A + ∠E = 180°

(∵ Interior angles on the same side of the transversal AE)

⇒ ∠E = 180° – ∠A

Since, BC = EC

∴ ∠E = ∠CBE = 180° – ∠A

Also, ∠ ABC =180° – ∠CBE (∵ ABE is straight line)

= 180°- 180°+ ∠A

⇒ ∠B = ∠A …(i)

**Now ∠A+ ∠D = 180°**

(ii)

(ii)

(∵ Interior angles on the same side of the transversal AD)

⇒ ∠D=180°-∠A

⇒ ∠D = 180° -∠B [From Eq. (i)]…(ii)

Also, ∠C+ ∠B = 180°

(∵ Interior angles on the same side of the transversal BC)

⇒ ∠C = 180°-∠B …..(iii)

From Eqs. (ii) and (iii), we get

∠C = ∠D

**Now, in ∆ABC and ∆ BAD, we have**

(iii)

(iii)

AB = BA (Common)

AD = BC (Given)

∠A=∠B [From Eq. (i)]

∴ ∆ABC ≅ ∆ BAD (By SAS)

**Since, ∆ABC ≅ ∆BAD**

(iv)

(iv)

∴ AC = BD

Hence proved.

**NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2**

**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 figure). AC is a diagonal. Show that**

**(i) SR || AC and SR = **12** AC**

**(ii) PQ = SR**

**(iii) PQRS is a parallelogram.**

**Solution:**

**Given:** P, Q, Ft and S are mid-points of the sides.

∴ AP = PB, BQ = CQ

CR = DR and AS = DS

**(i)** In ∆ADC, we have

S is mid-point of AD and R is mid-point of the DC.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.

∴ SB || AC …(i)

Also , SR = 12 AC …(ii)

**(ii)** Similarly, in ∆ABC, we have

PQ || AC ….(iii)

and PQ = 12 AC ….(iv)

Now, from Eqs. (i) and (iii), we get

SR = 12 AC …..(v)

**(iii)** Now, from Eqs. (i) and (iii), we get

PQ || SR

and from Eq. (v), PQ = SR

Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel.

So, PQRS is a parallelogram.

Hence proved.

**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.**

**Solution:**

**Given:** ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA

By mid-point theorem,

∴ PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.

∴ ∠EOF = 90°

Now, RQ || BD (By mid-point theorem)

⇒ RE || OF

Also, SP|| AC [From Eq. (i)]

⇒ FR || OE

∴ OERF is a parallelogram.

So, ∠ ERF = ∠EOF = 90°

(Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with ∠R = 90°

Hence, PQRS is a rectangle.

**Question 3.**

**ABCD is a rectangle and P, Q, R ans 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.

∴ ∠A = ∠B = ∠C= ∠D = 90°

and AD = BC, AB = CD

Also, given P, Q, R and S are mid-points of AB, BC, CD and DA .respectively.

∴ PQ || BD and PQ = 12 BD

In rectangle ABCD,

AC = BD

∴ PQ = SR …(ii)

Now, in ∆ASP and ∆BQP

AP = BP (Given)

AS = BQ (Given)

∠A = ∠B (Given)

∴ ∆ASP ≅ ∆BQP (By SAS)

∴ SP = PQ (By CPCT)…(ii)

Similarly, in ∆RDS and ∆RCQ,

SD = CQ (Given)

DR = RC (Given)

∠C=∠D (Given)

∴ ∆RDS ≅ ∆RCQ (By SAS)

∴ SR = RQ (By CPCT)…(iii)

From Eqs. (i), (ii) and (iii), it is clear that quadrilateral 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 figure). Show that F is the mid-point of BC.**

**Solution:**

**Given:** ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.

In ∆ABD, we have

EP\\AB

and E is mid-point of AD.

So, by theorem, if a line drawn through the mid-point of one side of a triangle parallel to another side bisect the third side.

∴ P is mid-point of BD.

Similarly, in ∆ BCD, we have,

PF || CD (Given)

and P is mid-point of BD.

So, by converse of mid-point theorem, F is mid-point of CB.

**Question 5.**

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

**Solution:**

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

**To prove:** Line segments AF and EC trisect the diagonal BD.

**Proof:** Since, ABCD is a parallelogram.

AB || DC

and AB = DC (Opposite sides of a parallelogram)

⇒ AE || FC and 12 AB = 12 DC

⇒ AF || FC and AF = FC

∴ AECF is a parallelogram.

∴ AF || FC

⇒ EQ || AP and FP || CQ

In ∆ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.

(By converse of mid-point theorem)

∴ BQ = PQ ….(i)

Again, in ∆DQC, F is the mid-point of DC and FP || CQ, so P is the mid-point of DQ. (By converse of mid-point theorem)

∴ QP = DP …(ii)

From Eqs. (i) and (ii), we get

BQ = PQ = PD

Hence, CE and AF 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.**

**Solution:**

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively, i.e., AS = SD, AP = BP, BQ = CQ and CR = DR. We have to show that PR and SQ bisect each other i.e., SO = OQ and PO = OR.

Now, in ∆ADC, S and R are mid-points of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. (By mid-point theorem)

∴ SR || AC and SR = 12 AC …(i)

Similarly, in ∆ ABC, P and Q are mid-points of AB and BC.

∴ PQ || AC and PQ = 12 AC (By mid-point theorem)…(ii)

From Eqs. (i) and (ii), we get

PQ || SR

and PQ = SR = 12 AC

∴ Quadrilateral PQRS is a parallelogram whose diagonals are SQ and PR. Also, we know that diagonals of a parallelogram bisect each other. So, SQ and PR 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) CM = MA = **12** AB**

**Solution:**

**Given:** ABC is a right angled triangle.

∠C = 90°

and M is the mid-point of AB.

Also, DM || BC

**(i)** In ∆ ABC, BC || MD and M is mid-point of AB.

∴ D is the mid-point of AC. (By converse of mid-point theorem)

**(ii)** Since, MD || BC and CD is transversal

∴ ∠ADM = ∠ACB (Corresponding angles)

But ∠ACB = 90°

∴ ∠ADM = 90° ⇒ MD ⊥ AC

**(iii)** Now, in ∆ ADM and ∆ CDM, we have

DM = MD (Common)

AD = CD (∵ D is mid point of AC)

∴ ∠ADM = ∠MDC (Each equal to 90°)

∴ ∆ ADM = ∆ CDM (By SAS)

∴ CM = AM (By CPCT)…(i)

Also, M is mid-point of AB.

∴ AM – BM = 12 AB ….(ii)

From Eqs. (i) and (ii), we get

CM = AM = 12 AB

Hence proved.