## RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A.

**Question 1.**

**Solution:**

In ∆ ABC, ∠A = 70° and AB = AC (given)

∴ ∠C = ∠B

(Angles opposite to equal sides)

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> 70° + ∠B + ∠B = 180°

(∴ ∠B = ∠C)

=> 2∠B = 180°- 70° = 110°

∠B = \(\frac { { 110 }^{ o } }{ 2 } \) = 55° and

Hence ∠B = 55°,∠C = 55° Ans.

**Question 2.**

**Solution:**

In ∆ ABC, ∠A= 100°

It is an isosceles triangle

∴AB = AC

∠B = ∠C

(Angles opposite to equal sides)

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> 100° + ∠B + ∠B = 180°

(∴ ∠B = ∠C)

=> 2∠B + 100° = 180°

=> 2∠B = 180°- 100° = 80°

∠B = \(\frac { { 80 }^{ o } }{ 2 } \) = 40°

and ∠C = 40°

∴ Base angles are 40°, 40° Ans.

**Question 3.**

**Solution:**

In ∆ ABC,

AB = AC

∴∠C = ∠B

(Angles opposite to equal sides)

But ∠B = 65°

∴ ∠ C = 65°

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> ∠A + 65° + 65° = 180°

=> ∠ A + 130° = 180°

=> ∠ A = 180° – 130°

=> ∠ A = 50°

Hence ∠ A = 50° and ∠ C = 65° Ans.

**Question 4.**

**Solution:**

In ∆ ABC

AB = AC

∴ ∠C = ∠B

(Angles opposite to equal sides)

But ∠A = 2(∠B + ∠C)

=> ∠B + ∠C = \(\frac { 1 }{ 2 } \) ∠A 2

But ∠A + ∠B + ∠C = 180°

(sum of angles of a triangle)

=> ∠A+ \(\frac { 1 }{ 2 } \) ∠A = 180°

=> \(\frac { 3 }{ 2 } \) ∠A = 180°

=> ∠A = 180° x \(\frac { 2 }{ 3 } \) = 120°

and ∠B + ∠C = \(\frac { 1 }{ 2 } \) ∠A = \(\frac { 1 }{ 2 } \) x 120°

= 60°

∴ ∠ B = ∠ C

∴ ∠ B = ∠ C = \(\frac { { 60 }^{ o } }{ 2 } \) = 30°

Hence ∠ A = 120°, ∠B = 30°, ∠C = 30° Ans.

**Question 5.**

**Solution:**

In ∆ ABC,

AB = BC and ∠B = 90°

∴ AB = BC

∴ ∠ C = ∠ A

(Angles opposite to equal sides)

Now ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> ∠ A + 90° + ∠ A = 180°

(∴ ∠C = ∠A)

=> 2∠A + 90° – 180°

=> 2∠ A = 180° – 90° = 90°

∠ A = \(\frac { { 90 }^{ o } }{ 2 } \) = 45°

∴ ∠ C = 45° (∴ ∠ C = ∠ A)

Hence, each of the equal angles is 45° Ans.

**Question 6.**

**Solution:**

**Given :** ∆ ABC is an isosceles triangle in which AB = AC

Base BC is produced to both sides upto D and E respectively forming exterior angles

∠ ABD and ∠ ACE

**To Prove :** ∠ABD = ∠ACE

**Proof :** In ∆ ABC

∴ AB = AC (given)

∴ ∠C = ∠B

(Angles opposite to equal sides)

=> ∠ ABC = ∠ACB

But ∠ ABC + ∠ABD = 180° (Linear pair)

Similarly ∠ACB + ∠ACE = 180°

∠ ABC + ∠ABD = ∠ACB + ∠ACE

But ∠ ABC = ∠ ACB (proved)

∠ABD = ∠ACE

Hence proved.

**Question 7.**

**Solution:**

∆ ABC is an equilateral triangle

∴ AB = BC = CA

and ∠A = ∠B = ∠C = 60°

The sides of the ∆ ABC are produced in order to D,E and F forming exterior angles

∠ACP, ∠BAE and ∠CBF

∆ACD + ∠ACB = 180°

=> ∠ACD + 60° = 180°

=> ∠ACD = 180° – 60°

=> ∠ACD = 120°

Similarly, ∠BAE + ∠BAC = 180°

=> ∠BAE + 60° = 180°

=> ∠BAE = 180° – 60° = 120°

and ∠BAF + ∠ABC = 180°

=> ∠BAF + 60° = 180°

=> ∠BAF = 180° – 60° = 120°

Hence each exterior angle of an equilateral triangle is 120°.

**Question 8.**

**Solution:**

**Given :** In the figure,

O is mid-point of AB and CD.

i.e. OA = OB and OC = OD

**To Prove :** AC = BD and AC || BD.

**Proof :** In ∆ OAC and ∆ OBD,

OA = OB {given}

OC = OD {given}

∠AOC = ∠BOD

(Vertically opposite angles)

∴∆ OAC ≅ ∆ OBD (S.A.S. axiom)

∴AC = BD (c.p.c.t.)

and ∠ C = ∠ D

But these are alt. angles

∴AC || BD

Hence proved.

**Question 9.**

**Solution:**

**Given :** In the figure,

PA ⊥ AB, QB ⊥ AB and

PA = QB, PQ intersects AB at O.

**To Prove :** O is mid-point of AB and PQ.

**Proof :** In ∆ AOP and ∆ BOQ,

∠ A = ∠ B

AP = BQ (given)

∠ AOP = ∠BOQ

(Vertically opposite angles)

∴ ∆AOP ≅ ∆ BOQ (A.A.S. axiom)

∴OA = OB (c.p.c.t)

and OP = OQ (c.p.c.t)

Hence, O is the mid-point of AB as well as PQ

**Question 10.**

**Solution:**

**Given :** Two line segments AB and CD intersect each other at O and OA = OB, OC = OD

AC and BD are joined.

**To Prove :** AC = BD

**Proof :** In ∆ AOC and ∆ BOD,

OA = OB {given}

OC = OD

∠AOC = ∠BOD

(vertically opposite angles)

∴ ∆ AOC ≅ ∆ BOD (S.A.S. axiom)

∴ AC = BD (c.p.c.t)

and ∠A = ∠D (c.p.c.t.)

Hence AC ≠ BD

Hence proved.

**Question 11.**

**Solution:**

**Given :** In the given figure,

l || m, m is mid-point of AB

CD is another line segment, which intersects AB at M.

**To Prove :** M is mid-point of CD

**Proof :** l || m

∴ ∠ CAM = ∠ MBD (Alternate angles)

Now, in ∆ AMC and ∆ BMD,

AM = MB (Given)

∠ CAM = ∠ MBD (proved)

∠ AMC = ∠BMD

(vertically opposite angles)

∆ AMC ≅ ∆ BMD (ASA axiom)

∴ CM = MD (c.p.c.t.)

Hence, M is mid-point of CD.

Similarly we can prove that M is mid point of any other line whose end points are on l and m.

Hence proved.

**Question 12.**

**Solution:**

**Given :** In ∆ ABC, AB = AC and in ∆ OBC,

OB = OC

**To Prove :** ∠ABO = ∠ACO

Construction. Join AO.

**Proof :** In ∆ ABO and ∆ ACO,

AB = AC (Given)

OB = OC (Given)

AO = AO (Common)

∴ ∆ ABO ≅ ∆ ACO (S.S.S. Axiom)

∴ ∠ABO = ∠ACO (c.p.c.t.)

Hence proved.

**Question 13.**

**Solution:**

**Given :** In ∆ ABC, AB = AC

D is a point on AB and a line DE || AB is drawn.

Which meets AC at E

**To Prove** : AD = AE

**Proof :** In ∆ ABC

AB = AC (given)

∴ ∠ C = ∠ B (Angles opposite to equal sides)

But DE || BC (given)

∴ ∠ D = ∠ B {corresponding angles}

and ∠ E = ∠ C

But ∠C = ∠B

∴ ∠E = ∠D

∴ AD = AE (Sides opposite to equal angles)

Hence proved.

**Question 14.**

**Solution:**

**Given :** In ∆ ABC,

AB = AC

X and Y are two points on AB and AC respectively such that AX = AY

**To Prove :** CX = BY

**Proof :** In ∆ AXC and ∆ AYB

AC = AB (given)

AX = AY (given)

∠ A = ∠ A (common)

∴ ∆ AXC ≅ ∆ AYB (S.A.S. axiom)

∴ CX = BY (c.p.c.t.)

Hence proved.

**Question 15.**

**Solution:**

**Given :** In the figure,

C is mid point of AB

∠DCA = ∠ECB and

∠DBC = ∠EAC

**To Prove :** DC = EC

**Proof :** ∠ DCA = ∠ FCB (given)

Adding ∠ DCE both sides,

∠DCA +∠DCE = ∠DCE + ∠ECB

=> ∠ACE = ∠ BCD

Now, in ∆ ACE and ∆ BCD,

AC = BC (C is mid-point of AB)

∠EAC = ∠DBC (given)

∠ACE =∠ BCD (proved)

∴ ∆ ACE ≅ ∆ BCD (ASA axiom)

∴ CE = CD (c.p.c.t.)

=> EC = DC

or DC = EC

Hence proved.

**Question 16.**

**Solution:**

**Given :** In figure,

BA ⊥ AC, DE ⊥ EF .

BA = DE and BF = DC

**To Prove :** AC = EF

**Proof :** BF = DC (given)

Adding FC both sides,

BF + FC = FC + CD

=> BC = FD.

Now, in right-angled ∆ ABC and ∆ DEF,

Hyp. BC = Hyp. FD (proved)

Side AB = side DE (given)

∴ ∆ ABC ≅ ∆DEF (RHS axiom)

∴ AC = EF (c.p.c.t.)

Hence proved.

**Question 17.**

**Solution:**

**To prove :** AE = CD

**Proof:** x° + ∠ BDC = 180° (Linear pair)

Similarly y°+ ∠AEB = 180°

∴ x° + ∠BDC = y° + AEB

But x° = y° (given)

∠ BDC = ∠ AEB

Now, in ∆ AEB and ∆ BCD,

AB = CB (given)

∠B = ∠B (common)

∠ AEB = ∠ BDC (proved)

∴ ∆ AEB ≅ ∆ BCD (AAS axiom)

∴ AE = CD. (c.p.c.t.)

Hence proved.

**Question 18.**

**Solution:**

**Given :** In ∆ ABC,

AB = AC.

Bisectors of ∠ B and ∠ C meet AC and AB in D and E respectively

**To Prove :** BD = CE

**Proof :** In ∆ ABC,

AB = AC

∠ B = ∠ C (Angles opposite to equal sides)

Now, in ∆ ABD and ∆ ACE,

AB = AC (given)

∠ A = ∠ A (common)

∠ ABD = ∠ACE

(Half of equal angles)

∴ ∆ A ABD ≅ ∆ ACE (ASA axiom)

∴ BD = CE (c.p.c.t)

Hence proved.

**Question 19.**

**Solution:**

**Given :** In ∆ ABC,

AD is median. BL and CM are perpendiculars on AD and AD is produced to E

**To prove :** BL = CM.

**Proof :** In ∆ BLD and ∆ CMD,

BD = DC (D is mid-point of BC)

∠LDB = ∠CDM

(Vertically opposite angles)

∠L = ∠M (each 90°)

∴ ∆ BLD ≅ ∆ CMD (A.A.S. axiom)

Hence, BL = CM (c.p.c.t)

Hence proved.

**Question 20.**

**Solution:**

**Given :** In ∆ ABC, D is mid-point of BC. DL ⊥ AB and DM ⊥ AC and DL = DM

**To prove :** AB = AC

**Proof:** In right angled ∆ BLD and ∆ CMD

Hyp. DL = DM (given)

Side BD = DC (D is mid-point of BC)

∴ ∆ BLD ≅ ∆ CMD (R.H.S. axiom)

∴ ∠B = ∠C (c.p.c.t.)

∴ AC = AB

(sides opposite to equal angles)

Hence AB = AC

Hence proved.

**Question 21.**

**Solution:**

**Given :** In ∆ AB = AC and bisectors of ∠B and ∠C meet at a point O. OA is joined.

**To Prove :** BO = CO and Ray AO is the bisector of ∠ A

**Proof :** AB = AC (given)

∴ ∠C = ∠B

(Angles opposite to equal sides)

=> \(\frac { 1 }{ 2 } \) ∠C = \(\frac { 1 }{ 2 } \) ∠B

=>∠OBC = ∠OCB

∴ in ∆OBC,

OB = OC (Sides opposite to equal angles)

Now in ∆ OAB and ∆ OCA

OB = OC (proved)

OA = OA (common)

AB = AC (given)

∴ ∆ OAB ≅ ∆ OCA (S.S.S. axiom)

∴ ∠OAB = ∠OAC (c.p.c.t.)

Hence OA is the bisector of ∠ A.

Hence proved.

**Question 22.**

**Solution:**

**Given :** ∆ PQR is an equilateral triangle and QRST is a square. PT and PS are joined.

**To Prove :** (i) PT = PS

(ii) ∠PSR = 15°

**Proof :** In ∆ PQT and ∆ PRS,

PQ = PR (Sides of equilateral ∆ PQR)

QT = RS (sides of in square PRST) .

∠PQT = ∠PRS

(each angle = 90° + 60° = 150°)

∴ ∆ PQT ≅ ∆ PRS (S.A.S. axiom)

∴ PT = PS (c.p.c.t)

In ∆ PRS, ∠PRS = 60° + 90° = 150°

∠ RPS + ∠ PSR = 180° – 150° = 30°

But ∠RPS = ∠PSR ( ∴PR = RS)

∴∠PSR + ∠PSR = 30°

=> 2∠PSR = 30°

∴ ∠PSR = \(\frac { { 30 }^{ o } }{ 2 } \) = 15°

Hence proved.

**Question 23.**

**Solution:**

**Given :** In right angle ∆ ABC, ∠B is right angle. BCDE is square on side BC and ACFG is also a square on AC.

AD and BF are joined.

**To Prove :** AD = BF

**Proof :** ∠ACF = ∠BCD (Each 90°)

Adding ∠ ACB both sides,

∠ ACF + ∠ ACB = ∠ BCD + ∠ ACB

=> ∠ BCF = ∠ ACD

Now in ∆ ACD and ∆ BCF,

AC = CF (sides of a squares)

CD = BC (sides of a square)

∴∠ ACD = ∠ BCF (proved)

∴ ∆ ACD ≅ ∆ BCF (S.A.S. axiom)

AD = BF (c.p.c.t)

Hence proved.

**Question 24.**

**Solution:**

Given : ∆ ABC is an isosceles in which AB = AC. and AD is the median which meets BC at D.

**To Prove :** AD is the bisector of ∠ A

**Proof :** In ∆ ABD and ∆ ACD,

AD = AD (Common)

AB = AC (Given)

BD = CD (D is mid-point of BC)

∴ ∆ ABD ≅ ∆ ACD (S.S.S. axiom)

∠ BAD = ∠ CAD (c.p.c.t.)

Hence, AD in the bisector of ∠ A.

**Question 25.**

**Solution:**

**Given :** ABCD is a quadrilateral in which AB || DC. P is mid-point of BC. AP and DC are produced to meet at Q.

**To Prove :** (i) AB = CQ.

(ii) DQ = DC + AB

**Proof:**

(i) In ∆ ABP and ∆ CPQ,

BP = PC (P is mid-point of BC)

∠ BAP = ∠ PQC (Alternate angles)

∠ APB = ∠ CPQ

(Vertically opposite angles)

∴ ∆ ABP ≅ ∆ CPQ (A.A.S. axiom)

∴ AB = CQ. (c.p.c.t.)

(ii) Now DQ = DC + CQ

=> DQ = DC + AB ( CQ = AB proved)

Hence proved.

**Question 26.**

**Solution:**

**Given :** In figure,

OA = OB and OP = OQ.

**To Prove :** (i) PX = QX (ii) AX = BX

**Proof :** In ∆OAQ and ∆OBP,

OA = OB. (Given)

OQ = OP (Given)

∠O = ∠O (Common)

∴ ∆ OAQ ≅ ∆ OBP (SAS axiom)

∴ ∠ A = ∠ B (c.p.c.t)

Now OA = OB and OP = OQ

Subtracting

OA – OP = OB – OQ

=>PA = QB

Now, in ∆ XPA and ∆ XQB,

PA = QB (Proved)

∠AXP = ∠BXQ

(Vertically opposite angles)

∠ A = ∠ B (Proved)

∴ ∆ XPA ≅ ∆ XQB (A.A.S. axiom)

∴ AX = BX (c.p.c.t.) and PX = QX (c.p.c.t.)

Hence proved.

**Question 27.**

**Solution:**

**Given :** ABCD is a square in which a point P is inside it. Such that PB = PD.

**To prove :** CPA is a straight line.

**Proof :** In ∆ APB and ∆ ADP,

AB = AD (Sides of a square)

AP = AP (Common)

PB = PD (Given)

∴ ∆ APB ≅ ∆ ADP (S.S.S. axiom)

∠APD = ∠ APB (c.p.c.t) …(i)

Similarly, in ∆ CBP and ∆ CPD,

CB = CD (Sides of a square)

CP – CP (Common)

PB = PD (Given)

∴ ∆ CBP ≅ ∆ CPD (S.S.S. axiom)

∴ ∠BPC = ∠CPD (c.p.c.t.) …(i)

Adding (i) and (ii),

∴ ∠ APD + ∠ CPD = ∠ APB + ∠ BPC

∠APC = ∠APC

∠APC = 180°

(v sum of angles at a point is 360°)

APC or CPA is a straight line.

Hence proved.

**Question 28.**

**Solution:**

**Given :**∆ ABC is an equilateral triangle PQ || AC and AC is produced to R such that CR = BP

**To prove :** QR bisects PC

**Proof :** ∴ PQ || AC

∴ ∠BPQ = ∠BCA

(Corresponding angles)

But ∠ BCA = 60°

(Each angle of the equilateral triangle)

and ∠ ABC or ∠QBP = 60°

∴ ∆ BPQ is also an equilateral triangle.

∴ BP = PQ

But BP = CR (Given)

∴ PQ = CR

Now in ∆ PQM and ∆ RMC,

PQ = CR (proved)

∠QMP = ∠RMC

(vertically opposite angles)

∠ PQM = ∠ MRC (alternate angles)

∴ ∆ PQM ≅ ∆ RMC (AAS axiom)

∴ PM = MC (c.p.c.t.)

Hence, QR bisects PC.

Hence proved

**Question 29.**

**Solution:**

**Given :** In quadrilateral ABCD,

AB = AD and BC = DC

AC and BD are joined.

**To prove :** (i) AC bisects ∠ A and ∠ C

(ii) AC is perpendicular bisector of BD.

**Proof :** In ∆ ABC and ∆ ADC,

AB = AD (given)

BC = DC (given)

AC = AC (common)

∴ ∆ ABC ≅ ∆ ADC (S.S.S. axiom)

∴ ∠BCA = ∠DCA (c.p.c.t.)

and ∠BCA = ∠DAC (c.p.c.t.)

Hence AC bisects ∠ A and ∠ C

(ii) In ∆ ABO and ∆ ADO,

AB = AD (given)

AO = AO (common)

∠BAO = ∠DAO

(proved that AC bisects ∠ A)

∴ ∆ ABO ≅ ∆ ADO (SAS axiom)

∴ BO = OD and ∠AOB = ∠AOD (c.p.c.t.)

But ∠ AOB + ∠ AOD = 180° (linear pair)

∠AOB = ∠AOD = 90°

Hence AC is perpendicular bisector of BD. Hence proved.

**Question 30.**

**Solution:**

**Given :** In ∆ ABC,

Bisectors of ∠ B and ∠ C meet at I

From I, IP ⊥ BC. IQ ⊥ AC and IR ⊥ AB.

IA is joined.

**To prove :** (i) IP = IQ = IR

(ii) IA bisects ∠ A.

**Proof :** (i) In ∆ BIP and ∆ BIR

BI = BI (Common)

and ∠P = ∠R (Each 90°)

∴ ∠ IBP ≅ ∠ IBR

( ∴ IB is bisector of ∠ B)

∴ ∆ BIP = ∆ BIR (A.A.S. axiom)

∴ IP = IR (c.p.c.t) …(i)

Similarly, in ∆ CIP and ∆ CIQ,

CI = CI (Common)

∠P = ∠Q (each = 90°) and ∠ICP = ∠ ICR

( IC is bisector of ∠ C)

∴ ∆CIP ≅ ∆CIQ (A.A.S. axiom)

∴ IP = IQ (c.p.c.t.) …(ii)

From (i) and (ii),

IP = IQ = IR.

(i) In right angled ∆ IRA and ∆ IQA,

Hyp. IA = IA (Common)

side IR = IQ (Proved)

∴ ∆ IRA ≅ ∆ IQA (R.H.S. axiom)

∴ ∠IAR = ∠ IRQ (c.p.c.t.)

Hence IA is the bisector of ∠ A

Hence proved.

**Question 31.**

**Solution:**

**Given.** P is a point in the interior of ∠ AOB

PL ⊥ OA and PM ⊥ OC are drawn and PL = PM.

**To prove :** OP is the bisector of ∠ AOB

**Proof:** In right angled ∆ OPL and ∆ OPM,

Hyp. OP OP (Common)

Side PL = PM (Given)

∴ ∆ OPL ≅ ∆ OPM (R.H.S. axiom)

∴ ∠POL = ∠POM (c.p.c.t.)

Hence, OP is the bisector of ∠AOB.

[ Hence proved.

**Question 32.**

**Solution:**

**Given :** ABCD is a square, M is midpoint of AB.

PQ is perpendicular to MC which meets CB produced at Q. CP is joined.

**To prove :** (i) PA = BQ

(ii) CP = AB + PA

**Proof :** (i) In ∆ PAM and ∆ QBM,

AM = MB (M is midpoint of AB)

∠AMP = ∠BMQ

(vertically opposite angles)

∠ PAM = ∠ QBM

(Each = 90° angles of a square)

∴ ∆ PAM ≅ ∆ QBM (A.S.A. axiom)

∴ AP = BQ

=> PA = BQ (c.p.c.t.)

and PM = QM (c.p.c.t.)

Now, in ∆ CPM and ∆ CQM,

CM = CM (Common)

PM = QM (Proved)

∠CMP =∠CMQ (Each 90° as PQ ⊥ MC)

∴ ∆ CPM ≅ ∆ CQM (SAS axiom)

∴ CP = CQ (c.p.c.t.)

= CB + BQ

= AB + PA

( CB = AB sides of squares and BQ = PA proved)

Hence, CP = AB + PA.

**Question 33.**

**Solution:**

**Construction**. Let AB be the breadth of the river. Mark any point M on the bank on which B is situated. Let O be the midpoint of BM. From M move along the path MN perpendicular to BM to a point N such that A, O, N are on the same straight line. Then MN is the required breadth of the river.

**Proof :** In ∆ ABO and ∆ MNO,

BO = OM (const.)

∠ AOB = ∠ MON (vertically opposite angle)

∠B = ∠M (each 90°)

∴ ∆ ABO ≅ ∆ MNO (A.S.A. axiom)

∴ AB = MN. (c.p.c.t.)

Hence, MN is the required breadth of the river.

**Question 34.**

**Solution:**

In ∆ ABC,

∠A = 36°, ∠B = 64°

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> 36° + 64° + ∠C = 180°

=> 100° + ∠C = 180°

=> ∠C = 180° – 100° = 80°

∴ ∠ C = 80° which is the greatest angle.

∴ The side AB, opposite to it is the longest side.

∴∠ A is the shortest angle

∴ BC is the shortest side.

**Question 35.**

**Solution:**

In ∆ ABC,

∴ ∠A = 90°

∴ ∠B + ∠C 180° – 90° = 90°

Hence, ∠ A is the greatest angle of the triangle.

Side BC, opposite to this angle be the longest side.

**Question 36.**

**Solution:**

In ∆ ABC,

∠ A = ∠ B = 45°

∴ ∠A + ∠B = 45° + 45° = 90°

and ∠C= 180°-90° = 90°

∴ ∠ C is the greatest angle.

∴ Side AB, opposite to ∠ C will be the longest side of the triangle.

**Question 37.**

**Solution:**

**Given :** In ∆ ABC, side AB is produced to D such that BD = BC.

∠B = 60°, ∠A = 70°

**To Prove :** (i) AD > CD (ii) AD > AC

**Proof :** In ∆ BCD,

Ext. ∠ B = 60°

∴ ∠ CBD = 180° – 60° = 120°

and ∠ BCD + ∠ BDC = Ext. ∠ CBA = 60°

But ∠BCD = ∠ BDC ( BC = BD)

∴ ∠BCD = ∠BDC = \(\frac { { 60 }^{ o } }{ 2 } \) = 30°.

∴ ∠ ACD = 50° + 30° = 80°

(i) Now in ∆ ACD,

∴ ∠ ACD > ∠ CAD ( 80° > 70°)

AD > CD .

(ii) and ACD > ∠D (80° > 30°)

∴ AD > AC

Hence proved.

**Question 38.**

**Solution:**

**Given :** In ∆ ABC, ∠B = 35°and ∠C = 65°

AX is the bisector of ∠ BA C meeting BC in X

∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> ∠ A + 35° + 65° = 180°

=> ∠A + 100° = 180°

=> ∠A = 180° – 100° = 80°

∴ AX is the bisector of ∠ BAC

∴ ∠ BAX = ∠ CAX = 40°

In ∆ ABX,

∠BAX = 40° and ∠B = 35°

∴ ∠BAX > ∠B

∴ BX > AX …(i)

and in ∆ AXC,

∠C = 65° and ∠CAX = 40°

∴ ∠C > ∠CAX

∴ AX > XC …(ii)

From (i) and (ii),

BX > AX > XC

or BX > AX > CX

**Question 39.**

**Solution:**

**Given :** In ∆ ABC,

AD is the bisector of ∠ A

**To prove :** (i) AB > BD and

(ii) AC > DC

**Proof :** (i) In ∆ ADC,

Ext. ∠ ADB > ∠ CAD

=>∠ ADB > ∠BAD ,

( AD is bisector of ∠ A)

In ∆ ABD,

AB > BD.

(ii) Again, in ∆ ADB,

Ext. ∠ADC > ∠BAD

=> ∠ADC > ∠CAD

( ∠ CAD = ∠BAD)

Now in ∆ ACD.

AC > DC.

Hence proved

**Question 40.**

**Solution:**

**Given :** In ∆ ABC, AB = AC

BC is produced to D and AD is joined.

**To Prove :** AD > AC

**Proof :** In ∆ ABC,

AB = AC (given)

∴ ∠B = ∠C (Angles opposite to equal sides)

Ext. ∠ ACD > ∠ ABC

∠ACB = ∠ABC

∴ ∠ABC > ∠ADC

Now, in ∆ ABD,

∴ ∠ABC > ∠ADC or ∠ADB

∴ AD > AC

Hence proved.

**Question 41.**

**Solution:**

**Given :** In ∆ ABC,

AC > AB and AD is the bisector of ∠ A which meets BC in D.

**To Prove :** ∠ADC > ∠ADB

**Proof :** In ∆ ABC,

AC > AB

∴∠B > ∠C

∴∠ 1 = ∠ 2 (AD is the bisector of ∠ A)

∴ ∠B + ∠2 = ∠C + ∠1

But in ∆ ADB

Ext. ∠ADC = ∠B + ∠2

and in ∆ ADC,

Ext. ∠ADB > ∠C + ∠1

∴ ∠B + ∠2 > ∠C + ∠1 (Proved)

∴ ∠ ADC > ∠ ADB Hence proved.

**Question 42.**

**Solution:**

**Given :** In ∆ PQR,

S is any point on QR and PS is joined

**To Prove :** PQ + QR + RP > 2PS

**Proof :** In ∆ PQS,

PQ + QS > PS

(Sum of two sides of a triangle is greater than the third side) …(i)

Similarly, in ∆ PRS.

PQ + SR > PS …(ii)

Adding (i), and (ii),

PQ + QS + PR + SR > PS + PS

=> PQ + QS + SR + PR > 2PS

=> PQ + QR + RP > 2PS

Hence proved.

**Question 43.**

**Solution:**

**Given :** O is the centre of the circle and XOY is its diameter.

XZ is the chord of this circle

**To Prove :** XY > XZ

Const. Join OZ

**Proof :** OX, OZ and OY are the radii of the circle

∴ OX = OZ = OY

In ∆ XOZ,

OX + OZ > XZ (Sum of two sides of a triangle is greater than its third side)

=> OX + OY > XZ (∴ OZ = OY)

=> OXY > XZ

Hence proved.

**Question 44.**

**Solution:**

**Given :** In ∆ ABC, O is any point within it OA, OB and OC are joined.

**To Prove :** (i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > \(\frac { 1 }{ 2 } \)

(AB + BC + CA)

Const. Produce BO to meet AC in D.

**Proof :** (i) In ∆ ABD,

AB + AD > BD …(i)

(Sum of two sides of a triangle is greater than its third side)

=> AB + AD > BO + OD …(i)

Similarly in ∆ ODC

OD + DC > OC. …(ii)

Adding (i) and (ii)

AB + AD + OD + DC > OB + OD + OC

=> AB + AD + DC > OB + OC

AB + AC > OB + OC.

(ii) In (i) we have proved that

AB + AC > OB + OC

Similarly, we can prove that

AC + BC > OC + OA

and BC + AB > OA + OB

Adding, we get:

AB + AC + AC + BC + B + AB > OB + OC + OC + OA + OA + OB

=> 2(AB + BC + CA) > 2(OA + OB + OC)

=> AB + BC + CA > OA + OB + OC.

(iii) In ∆ AOB,

OA + OB > AB

Similarly, in ∆ BOC

OB + OC > BC

and in ∆ COA

OC + OA > CA

adding we get:

OA + OB + OB + OC + OC + OA > AB + BC + CA

=> 2(OA + OB + OC) > (AB + BC + CA)

=> OA + OB + OC > \(\frac { 1 }{ 2 } \) (AB + BC + CA)

Hence proved.

**Question 45.**

**Solution:**

Sides of ∆ ABC are AB = 3cm, BC = 3.5cm and CA = 6.5cm

We know that sum of any two sides of a triangle is greater than its third side.

Here, AB = 3 cm and BC = 3.5 cm

∴ AB + BC = 3cm + 3.5 cm = 6.5 cm and CA = 6.5 cm

∴ AB + BC = CA

Which is not possible to draw the triangle.

Hence, we cannot draw the triangle with the given data.

Hope given RS Aggarwal Class 9 Solutions Chapter 5 Congruence of Triangles and Inequalities in a Triangle Ex 5A are helpful to complete your math homework.

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