## 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.**

**In a ∆ ABC, if AB = AC and ∠ A = 70°, find ∠B and ∠C.**

**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 = = 55° and

Hence ∠B = 55°,∠C = 55° Ans.

**Question 2.**

**The vertical angle of an isosceles triangle is 100°. Find its base angles.**

**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 = = 40°

and ∠C = 40°

∴ Base angles are 40°, 40° Ans.

**Question 3.**

**In a ∆ ABC, if AB = AC and ∠B = 65° find ∠ C and ∠ A.**

**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.**

**In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.**

**Solution:**

In ∆ ABC

AB = AC

∴ ∠C = ∠B

(Angles opposite to equal sides)

But ∠A = 2(∠B + ∠C)

=> ∠B + ∠C = ∠A 2

But ∠A + ∠B + ∠C = 180°

(sum of angles of a triangle)

=> ∠A+ ∠A = 180°

=> ∠A = 180°

=> ∠A = 180° x = 120°

and ∠B + ∠C = ∠A = x 120°

= 60°

∴ ∠ B = ∠ C

∴ ∠ B = ∠ C = = 30°

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

**Question 5.**

**What is the measure of each of the equal angles of a right-angled isosceles triangle ?**

**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 = = 45°

∴ ∠ C = 45° (∴ ∠ C = ∠ A)

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

**Question 6.**

**If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.**

**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.**

**Find the measure of each exterior angle of an equilateral triangle.**

**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.**

**In the given figure, O is the midpoint of each of the line segments AB and CD. Prove that AC = BD and AC || BD.**

**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.**

**In the adjoining figure, PA ⊥ AB, QB ⊥ AB and PA = QB. If PQ intersects AB at O, show that O is the midpoint of AB as well as that of PQ.**

**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.**

**Let the line segments AB and CD intersect at O in such a way that OA = OD and OB = OC. Prove that AC = BD but AC may not be parallel to BD.**

**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.**

**In the given figure, l || m and M is the midpoint of AB. Prove that M is also the mid-point of any line segment CD having its end points at l and m respectively**

**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.**

**In the given figure, AB = AC and OB = OC. Prove that ∠ABO = ∠ACO.**

**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.**

**In the given figure, ABC is a triangle in which AB = AC and D is a point on AB. Through D, a line DE is drawn parallel to BC and meeting AC at E. Prove that AD = AE.**

**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.**

**In the adjoining figure, X and Y are respectively two points on equal sides AB and AC of ∆ ABC such that AX = AY. Prove that CX = BY.**

**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.**

**In the given figure, C is the mid-point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠ EAC, prove that DC = EC.**

**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.**

**In the given figure, BA ⊥ AC and DE ⊥ EF such that BA = DE and BF = DC. Prove that AC = EF.**

**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.**

**In the given figure, if x = y and AB = CB, then prove that AE = CD.**

**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.**

**ABC is a triangle in which AB = AC. If the bisectors of ∠ B and ∠ C meet AC and AB in D and E respectively, prove that BD = CE.**

**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.**

**In the adjoining figure, AD is a median of ∆ ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.**

**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.**

**In ∆ ABC, D is the midpoint of BC. If DL ⊥ AB and DM ⊥ AC such that DL = DM, prove that AB = AC.**

**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.**

**In ∆ ABC, AB = AC and the bisectors of ∠ B and ∠ C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠ A.**

**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)

=> ∠C = ∠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.**

**In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that (i) PT = PS, (ii) ∠ PSR = 15°.**

**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 = = 15°

Hence proved.

**Question 23.**

**In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.**

**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.**

**Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.**

**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.**

**In the given figure, ABCD is a quadrilateral in which AB || DC and P is the midpoint of BC. On producing, AP and DC meet at Q. Prove that (i) AB = CQ, (ii) DQ = DC + AB.**

**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.**

**In the given figure, OA = OB and OP = OQ. Prove that (i) PX = QX, (ii) AX = BX.**

**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.**

**In the given figure, ABCD is a square and P is a point inside it such that PB = PD. Prove that CPA is a straight line.**

**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.**

**In the given figure, ABC is an equilateral triangle ; PQ || AC and AC is produced to R such that CR = BP. Prove that QR bisects PC.**

**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.**

**In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that (i) AC bisects ∠ A and ∠ C, (ii) AC is the perpendicular bisector of BD.**

**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.**

**In the given figure, the bisectors of ∠B and ∠C of ∆ABC meet at I. If IP ⊥ BC, IQ⊥ CA and IR ⊥ AB, prove that (i) IP = IQ = IR, (ii) IA bisects ∠A.**

**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.**

**In the adjoining figure, P is a point in the interior of ∠ AOB. If PL ⊥ OA and PM ⊥ OB such that PL = PM, show that OP is the bisector of ∠ AOB.**

**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.**

**In the given figure, ABCD is a square, M is the midpoint of AB and PQ ⊥ CM meets AD at P and CB produced at Q. Prove (i) PA = BQ (ii) CP = AB + PA**

**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.**

**In the adjoining figure, explain how one can find the breadth of the river without crossing it.**

**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.**

**In ∆ ABC, if ∠ A = 36° and ∠B = 64°, name the longest and the shortest sides of the triangle.**

**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.**

**In ∆ ABC, if ∠ A = 90°, which is the longest side.**

**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.**

**In ∆ ABC, if ∠ A = ∠ B = 45°, name the longest side**

**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.**

**In ∆ ABC, side AB is produced to D such that BD = BC. If ∠B = 60° and ∠ A = 70°, prove that (i) AD > CD and (ii) AD > AC**

**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 = = 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.**

**In ∆ ABC, ∠ B = 35°, ∠ C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.**

**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.**

**In ∆ ABC, if AD is the bisector of ∠ A, show that AB > BD and AC < DC.**

**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.**

**In the given figure, ABC is a triangle in which AB = AC. If D be a point on BC produced, prove that AD > AC.**

**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.**

**In the adjoining figure, AC > AB and AD is the bisector of ∠ A. Show that ∠ADC > ∠ADB.**

**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.**

**In ∆ PQR, if S is any point on the side QR, show that PQ + QR + RP > 2PS.**

**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.**

**In the given figure, O is the centre of the circle and XOY is a diameter. If XZ is any other chord of the circle, show that XY > XZ**

**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.**

**If O is a point within ∆ ABC, show that:**

**(i) AB + AC > OB + OC**

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

**(iii) OA + OB + OC > **

**(AB + BC + CA)**

**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 >

(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 > (AB + BC + CA)

Hence proved.

**Question 45.**

**Can we draw a triangle ABC with AB 3 cm, BC = 3.5 cm and CA = 6.5 cm ? Why ?**

**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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Molisha singh says

July 24, 2018 at 7:43 pmThese solutions have been very helpful for me.