Read RD Sharma Solutions Class 9 Chapter 10 Congruent Triangles below, students should study RD Sharma class 9 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 9 Mathematics have been prepared by teacher of Grade 9. These RD Sharma class 9 Solutions have been designed as per the latest NCERT syllabus for class 9 and if practiced thoroughly can help you to score good marks in standard 9 Mathematics class tests and examinations

**Exercise 10.1**

**Question 1: In figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that segment DE ∥ BC.**

**Solution:**

In ∆ABC and ∆ADE

It is given that,

BA = AD (Given)

CA = AE (Given)

∠BAC = ∠DAE (Vertically opposite angle)

∆ABC ≅ ∆ADE (By SAS congruent criterion)

BC = DE (CPCT)

∠C = ∠E and ∠B = ∠D (alternate interior angle)

BC || ED

Hence Prove.

**Question 2: In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.**

**Solution:**

Draw a figure based on given instruction,

In △PQR,

It is given that,

PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP.

∠QPR = ∠QRP (two sides of the triangle are equal, so △PQR is an isosceles triangle)

L is the midpoint of PQ and M is midpoints QR.

PL = LQ = QM = MR

Now, consider Δ LPN and Δ MRN,

LP = MR (Proved above)

∠LPN = ∠MRN (∠QPR = ∠LPN and ∠QRP = ∠MRN)

PN = NR (N is midpoint of PR)

Δ LPN ≅ Δ MRN (By SAS congruence criterion)

LN = MN (CPCT)

Hence Proved.

**Question 3: In figure, PQRS is a square and SRT is an equilateral triangle. Prove that**

**(i) PT = QT (ii) ∠ TQR = 15°**

**Solution:**

(i) It is given that, PQRS is a square and SRT is an equilateral triangle.

PQRS is a square

PQ = QR = RS = SP_________(i)

∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90° (Property of a square)

△SRT is an equilateral triangle

SR = RT = TS_________(ii)

∠ TSR = ∠ SRT = ∠ RTS = 60° (Property of equilateral triangle)

From (i) and (ii) we get,

PQ = QR = SP = SR = RT = TS_________(iii)

From figure we get,

∠TSP = ∠TSR + ∠ RSP

∠TSP = 60° + 90°

∠TSP = 150°

∠TRQ = ∠TRS + ∠ SRQ

∠TRQ = 60° + 90°

∠TRQ = 150°

∠TSP = ∠TRQ = 150°__________(iv)

From ΔTSP and ΔTRQ we get,

TS = TR (Proved above)

SP = RQ (Proved above)

∠TSP = ∠TRQ (each 150°)

ΔTSP ≅ ΔTRQ (SAS congruence criterion)

PT = QT (Corresponding parts of congruent triangles)

Hence proved

(ii) In Δ TQR.

QR = TR

ΔTQR is an isosceles triangle.

∠QTR = ∠TQR (angles opposite to equal sides)

∠QTR + ∠ TQR + ∠TRQ = 180° (Sum of angles in a triangle is 180°)

∠TQR + ∠ TQR + ∠TRQ = 180°

2∠TQR + 150° = 180°

2∠TQR = 30°

∠TQR = 15°

Hence proved

**Question 4: Prove that the medians of an equilateral triangle are equal.**

**Solution:**

It is given that, △ABC is an equilateral and Let D, E, F are midpoints of BC, CA and AB.

Here, AD, BE and CF are medians of △ABC.

D is midpoint of BC

BD = DC

F is midpoint of AB

CE = EA

E is midpoint of AC

AF = FB

ΔABC is an equilateral triangle

AB = BC = CA (equilateral triangle)

BD = DC = CE = EA = AF = FB

∠ABC = ∠BCA = ∠CAB = 60° (angles in a equilateral triangle are 60°)

In ΔABD and ΔBCE

AB = BC (sides of a equilateral triangle)

BD = CE (sides of a equilateral triangle)

∠ABD = ∠BCE (each 60°)

ΔABD ≅ ΔBCE (By SAS congruence criterion)

AD = BE (Corresponding parts of congruent triangles) __________(1)

In ΔBCE and ΔCAF,

BC = CA (sides of a equilateral triangle)

∠BCE = ∠CAF (each 60°)

CE = AF (Proved above)

ΔBCE ≅ ΔCAF (By SAS congruence criterion)

BE = CF (CPCT) ____________(2)

From equation (1) and (2), we get

AD = BE = CF (Median of equilateral triangle)

The medians of an equilateral triangle are equal.

Hence proved

**Question 5: In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.**

**Solution:**

ΔABC is an isosceles triangle since AB = AC

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

∠ A + ∠ B + ∠ C = 180° (Sum of angles in a triangle is 180°)

∠ A + ∠ B + ∠ B = 180°

120° + 2∠B = 180°

2∠B = 180° – 120°

2∠B = 60°

∠ B = 30°

Hence, the value of ∠B = ∠C = 30°.

**Question 6: In a Δ ABC, if AB = AC and ∠ B = 70°, find ∠ A.**

**Solution:**

It is given that, in ΔABC, AB = AC and ∠B = 70°

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

∠ B = ∠ C = 70°

∠ A + ∠ B + ∠ C = 180° (Sum of angles in a triangle is 180°)

∠ A + 70° + 70° = 180°

∠ A = 180° – 140°

∠ A = 40°

Hence, the value of ∠A is 40°.

**Question 7: The vertical angle of an isosceles triangle is 100°. Find its base angles.**

**Solution:**

It is given that ΔABC is an isosceles triangle.

AB = AC

Given that vertical angle A is 100°

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

∠A + ∠B + ∠C = 180° (Sum of interior angles of a triangle is 180°)

100° + ∠B + ∠B = 180°

2∠B = 180° - 100°

∠B = (80°)/2

∠B = 40°

Hence, the value of base angles ∠B and ∠C is 40°.

**Question 8: In a figure AB = AC and ∠ACD = 105°. Find ∠BAC.**

**Solution:**

It is given that,

AB = AC and ∠ACD = 105°

BCD is a Straight line.

∠BCA + ∠ACD = 180°

∠BCA + 105° = 180°

∠BCA = 180° - 105°

∠BCA = 75°

ΔABC is an isosceles triangle.

AB = AC

∠ABC = ∠ ACB (Angles opposite to equal sides are equal)

∠ACB = 75°

∠ABC = ∠ACB = 75°

∠ABC + ∠BCA + ∠CAB = 180° (Sum of Interior angles of a triangle is 180°)

75° + 75° + ∠CAB =180°

150° + ∠BAC = 180°

∠BAC = 180° - 150°

∠BAC = 30°

Hence, the value of ∠BAC is 30°.

**Question: 9 Find the measure of each exterior angle of an equilateral triangle.**

**Solution:**

It is given that, ∆ABC equilateral triangle.

AB = BC = CA

∠ABC = ∠BCA = CAB

∠ABC = ∠BCA = CAB = 60° (all angles in a equilateral triangle are equal)

Extend side BC to D,

CA to E

AB to F

BCD is a straight line.

∠BCA + ∠ACD = 180° (Angles made in on it are 180°)

60° + ∠ACD = 180°

∠ACD = 120°

The exterior angle of an equilateral triangle is 120°

Hence proved

**Question: 10 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:**

AB = AC (Sides of a isosceles triangle)

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

Exterior angles ∠ACD

∠A + ∠B = ∠ACD _______(1) (Sum of opposite interiors angles)

Exterior angles ∠ACD

∠A + ∠C = ∠ABE _______(2) (Sum of opposite interiors angles)

By equation (1) and (2)

∠A + ∠B = ∠A + ∠C

∠ACD = ∠ABE

Hence proved

**Question 11: In Figure AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.**

**Solution:**

In ∆ABC

AB = AC

∠ABC = ∠ACB (Angles opposite to equal sides are equal)

In ∆DBC

DB = DC

∠DBC = ∠DCB (Angles opposite to equal sides are equal)

∠ABC = ∠ABD + ∠DBC

∠ABC - ∠DBC = ∠ABD

∠ACB = ∠ACD = ∠DCB

∠ACB - ∠DCB = ∠ACD

(∠ABC)/(∠ACD)=(∠ABC-∠DBC)/(∠ACB-∠DCB)

(∠ABC)/(∠ACD)=(∠ABC-∠DBC)/(∠ABC-∠DBC)

(∠ABC)/(∠ACD)=1/1

∠ABC∶ ∠ACD=1:1

**Question: 12 Determine the measure of each of the equal angles of a right-angled isosceles triangle.**

**OR**

**ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.**

**Solution:**

ABC is a right angled triangle.

Consider on a right - angled isosceles triangle ABC such that

∠A = 90°

AB = AC

∠C = ∠B _______(i) (Angles opposite to equal sides are equal)

∠A + ∠B + ∠C =180° (Sum of angles in a triangle is 180°)

90° + ∠ B+ ∠ B = 180°

2∠B = 90°

∠B = 45°

∠B = 45°

∠C = 45°

Hence, the angles of a right-angled isosceles triangle is 45°

**Question: 13 AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. (10.26). Show that the line PQ is perpendicular bisector of AB.**

**Solution:**

It is given that,

AB is line segment and P, Q are points.

AP = BP _________(i)

AQ = BQ ________(ii)

We have to prove that PQ is perpendicular bisector of AB.

In ΔPAQ and ΔPBQ,

AP = BP (Proved above)

AQ = BQ (Proved above)

PQ - PQ (Common)

Δ PAQ ≅ Δ PBQ (by SSS congruence) ________(iii)

∠APC = ∠BPC (CPCT)

Now, we can observe that ∆APB and ∆ABQ are isosceles triangles. [From (i) and (ii)]

∠PAB = ∠ABQ (Angles opposite to equal sides are equal)

∠QAB = ∠QBA (Angles opposite to equal sides are equal)

In ΔPAC and ΔPBC

AP = BP (Proved above)

∠APC = ∠BPC (Proved above by equation 3)

PC = PC (common side)

ΔPAC ≅ ΔPBC (SAS congruency of triangle)

AC = CB (CPCT)

∠PCA = ∠PBC (CPCT)

ACB is Straight line

∠ACP + ∠ BCP = 180°

∠ACP = ∠PCB

∠ACP = ∠PCB = 90°

AC = CB, C is the midpoint of AB

PC is the perpendicular bisector of AB

Hence, C is a point on the line PQ, PQ is the perpendicular bisector of AB.

**Exercise 10.2**

**Question 1: In figure, it is given that RT = TS, ∠ 1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT.**

**Solution:**

It is given that,

RT = TS

∠1 = 2 ∠2

∠4 = 2∠3

Need to prove: ΔRBT ≅ ΔSAT

RB and SA intersect each other at point O.

∠AOR = ∠BOS (Vertically opposite angles)

∠1 = ∠4

2∠2 = 2∠3 (Given)

∠2 = ∠3

In ΔTRS,

RT = TS

ΔTRS is an isosceles triangle

∠TRS = ∠TSR (Angles opposite to equal sides are equal)

∠TRS = ∠TRB + ∠2_______(1)

∠ TSR = ∠TSA + ∠3_______(2)

From equation (1) and (2) we get

∠TRB + ∠2 = ∠TSA + ∠3

∠TRB = ∠TSA (∠2 = ∠3)

In ΔRBT and ΔSAT

RT = ST (Given)

∠TRB = ∠TSA (Proved above)

∠T = ∠T (common)

Δ RBT ≅ Δ SAT (ASA criterion of congruence)

Hence Prove.

**Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.**

**Solution:**

It is given that, BC ∥ AD and BC = AD

Need to prove: AB and CD bisect at O.

In ΔAOD and ΔBOC

∠OCB =∠ODA (alternate interior angles)

AD = BC (given)

∠OBC = ∠OAD (alternate interior angles)

Δ AOD ≅ Δ BOC (ASA Criterion)

OA = OB (By CPCT)

OD = OC (By CPCT)

AB and CD bisect each other at O.

Hence Proved.

**Question 3: BD and CE are bisectors of ∠B and ∠C of an isosceles Δ ABC with AB = AC. Prove that BD = CE.**

**Solution:**

It is given that,

AB = AC

∠ABC = ∠ACB________(i) (Angles opposite to equal sides are equal)

BD is bisector of ∠B

∠ABD = ∠DBC

CE is bisectors of ∠C

∠BCE = ∠ECA

In ΔEBC and ΔDCB we get,

∠EBC = ∠DCB (Proved above)

BC = BC (Common)

∠BCE = ∠CBD (Proved above)

ΔEBC ≅ ΔDCB (By ASA congruence criterion)

CE = BD (CPCT)

BD = CE (CPCT)

Hence proved

**Exercise 10.3**

**Question 1: In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.**

**Solution:**

It is given that, two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other.

In right triangle △ABC and △DEF, we get

∠B = ∠E (each 90°)

AB = DE (give)

∠C = ∠F (give)

ΔABC ≅ ΔDEF (By AAS congruence criterion)

Both the triangles are congruent.

Hence proved

**Question 2: If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.**

**Solution:**

Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle, ∠EAC and AD ∥ BC.

From figure,

∠1 = ∠2 (as AD is a bisector of ∠EAC) _______(i)

∠2 = ∠4 (alternative angle) _______(ii)

∠1 = ∠3 (corresponding angles) _______(iii)

From equation (i), (ii) and (iii), we get

∠3 = ∠4

AB = AC

ΔABC is an isosceles triangle.

**Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.**

**Solution:**

It is given that, ΔABC be isosceles.

AB = AC

∠B = ∠C

Vertex angle A is twice the sum of the base angles B and C.

∠A = 2(∠B + ∠C)

∠A = 2(∠B + ∠B)

∠A = 2(2∠B)

∠A = 4∠B

∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)

4∠B + ∠B + ∠B = 180°

6∠B =180°

∠B = 30°

∠B = ∠C

∠B = ∠C = 30°

∠A = 4∠B

∠A = 4 x 30°

∠A = 120°

Hence, the value of angles of the given triangles are 30° and 30° and 120°.

**Question 4: PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.**

**Solution:**

It is given that, PQR is a triangle such that PQ = PR and S is any point on the side PQ and ST ∥ QR.

Here,

PQ = PR

So, △PQR is an isosceles triangle.

∠PQR = ∠PRQ

∠PST = ∠PQR (Corresponding angles)

∠PTS = ∠PRQ (Corresponding angles)

As, ∠PQR = ∠PRQ

∠PST = ∠PTS

In Δ PST,

∠ PST = ∠ PTS

Δ PST is an isosceles triangle.

Hence, PS = PT.

Hence proved

**Exercise 10.4**

**Question 1: In figure, It is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.**

**Solution:**

It is given that,

AB = CD and AD = BC

In ΔADC and ΔCBA

AB = CD (Given)

BC = AD (Given)

And AC = AC (Common)

ΔADC ≅ ΔCBA (by SSS congruence criterion)

Hence proved

**Question 2: In a Δ PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.**

**Solution:**

It is given that, in ΔPQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively.

Join L to M, M to N, N to L

PL = LQ, QM = MR and RN = NP (As L, M and N are mid-points of PQ, QR and RP)

PQ = QR

PL = LQ = QM = MR = PN = LR ________(i) (by mid-point theorem)

MN ∥ PQ

MN = PQ/2

MN = PL = LQ _______(ii)

LN ∥ QR

LN = QR/2

LN = QM = MR _______(iii)

From equation (i), (ii) and (iii), we get

PL = LQ = QM = MR = MN = LN

LN = MN

Hence Proved

**Exercise 10.5**

**Question 1: ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.**

**Solution:**

It is given that D is the midpoint of BC.

PD = DQ in a triangle ABC.

In △BDP and △CDQ

PD = QD (Given)

BD = DC (D is mid-point)

∠BPD = ∠CQD (each 90°)

△BDP ≅ △CDQ (By RHS Criterion)

BP = CQ (CPCT) ________(i)

In △APD and △AQD

PD = QD (given)

AD = AD (common)

APD = AQD (each 90°)

△APD ≅ △AQD (By RHS Criterion)

PA = QA (CPCT) ________(ii)

Adding (i) and (ii)

BP + PA = CQ + QA

AB = AC

Hence it is proved that two sides of the triangle are equal, so ABC is an isosceles.

**Question 2: ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that Δ ABC is isosceles**

**Solution:**

ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively s.t. BE = CF.

In Δ BCF and Δ CBE,

∠BFC = ∠CEB (each 90°)

BC = CB [Common side]

And CF = BE (Given)

ΔBFC ≅ ΔCEB (By RHS congruence criterion)

∠ FBC = ∠ EBC (CPCT)

∠ABC = ∠ACB (CPCT)

AC = AB (Opposite sides to equal angles are equal in a triangle)

Two sides of triangle ABC are equal.

Hence, ΔABC is isosceles.

**Question 3: If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.**

**Solution:**

It is given that, an angle ABC and BP is the mid-point of angle B.

Construction: Draw perpendicular PN and PM on the sides BC and BA.

In Δ BPM and Δ BPN

∠ BMP = ∠ BNP (each 90°)

BP = BP (Common)

MP = NP (given)

ΔBPM ≅ ΔBPN (By RHS congruence criterion)

∠MBP = ∠NBP (CPCT)

BP is the angular bisector of ∠ABC.

Hence proved

**Question 4: In figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.**

**AD ⊥ CD and CB ⊥ CD.**

**Solution:**

It is given that, in the figure, AD ⊥ CD and CB ⊥ CD.

AQ = BP

DP = CQ ______(i)

PQ added on both sides of equation (i)

DP + PQ = PQ + QC

DQ = PC ... (i)

In ∆DAQ and ∆CBP

∠ADQ = ∠BCP (each 90°)

AQ = BP (given)

DQ = PC (proved above)

ΔDAQ ≅ ΔCBP ((by RHS congruence criterion)

∠DAQ = ∠CBP (CPCT)

Hence proved

**Question 5: ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.**

**Solution:**

It is given that ABCD is a square, X and Y are points on sides AD and BC respectively and AY = BX.

Join B and X, A and Y.

ABCD is a square,

∠DAB = ∠CBA (each 90°)

∠XAB = ∠YAB (each 90°)

In ∆XAB and ∆YBA

∠XAB = ∠YBA (each 90°)

BX = AY (given)

And AB = BA (Common)

ΔXAB ≅ ΔYBA (by RHS congruence criterion)

BY = AX (CPCT)

∠BAY = ∠ABX (CPCT)

Hence proved

**Question 6: Which of the following statements are true (T) and which are false (F):**

**(i) Sides opposite to equal angles of a triangle may be unequal.**

**(ii) Angles opposite to equal sides of a triangle are equal**

**(iii) The measure of each angle of an equilateral triangle is 60**

**(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.**

**(v) The bisectors of two equal angles of a triangle are equal.**

**(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.**

**(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.**

**(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.**

**(ix) Two right-angled triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.**

**Solution:**

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) False

(vii) False

(viii) False

(ix) True

**Question 7: Fill the blanks.**

**In the following so that each of the following statements is true.**

**(i) Sides opposite to equal angles of a triangle are _________**

**(ii) Angle opposite to equal sides of a triangle are _________**

**(iii) In an equilateral triangle all angles are _________**

**(iv) In ΔABC, if ∠A = ∠C, then AB = _________**

**(v) If altitudes CE and BF of a triangle ABC are equal, then AB _________**

**(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is _________ CE.**

**(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ΔABC ≅ Δ _________**

**Solution:**

(i) Sides opposite to equal angles of a triangle are equal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) In an equilateral triangle all angles are equal.

(iv) In a ΔABC, if ∠A = ∠C, then AB = BC.

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC

(vi) In an isosceles triangle ΔABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE

(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then, ΔABC = ΔEFD.

**Exercise 10.6**

**Question 1: In Δ ABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.**

**Solution:**

It is given that, ΔABC, ∠A = 40° and ∠B = 60°

∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)

40° + 60° + ∠C = 180°

∠C = 180° – 100° = 80°

∠C = 80°

40° < 60° < 80°

∠A < ∠B < ∠C

∠C is greater angle and ∠ A is smaller angle.

∠A < ∠B < ∠C (side opposite to greater angle is larger and side opposite to smaller angle is smaller)

BC < AC < AB

Hence we can say that, AB is longest and BC is shortest side.

**Question 2: In a Δ ABC, if ∠ B = ∠ C = 45°, which is the longest side?**

**Solution:**

It is given that, ΔABC, ∠B = ∠C = 45°

∠A + ∠B + ∠C = 180° (Sum of angles in a triangle is 180°)

∠A + 90° = 180°

∠A = 180° – 90°

∠A = 90°

∠B = ∠C < ∠A

Hence, BC is the longest side.

**Question 3: In Δ ABC, side AB is produced to D so that BD = BC. If ∠ B = 60° and ∠ A = 70°.**

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

**Solution:**

It is given that, Δ ABC, side AB is produced to D so that BD = BC ∠ B = 60°, and ∠ A = 70°

Construction: Join C and D

∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)

70° + 60° + ∠C = 180°

130° + ∠C = 180°

∠C = 180° – 130°

∠C = 50°

∠ACB = 50° ____________(i)

In ΔBDC

∠DBC = 180° – ∠ ABC

∠DBC = 180 – 60°

∠DBC = 120°

DBA is a straight line

BD = BC (given)

∠BCD = ∠BDC (angles opposite to equal sides are equal)

∠DBC + ∠BCD + ∠BDC = 180° (Sum of angles in a triangle is 180°)

120° + ∠ BCD + ∠ BCD = 180°

120° + 2∠BCD = 180°

2∠BCD = 180° – 120°

2∠BCD = 60°

∠BCD = 30°

∠BCD = ∠BDC = 30° ____________(ii)

In Δ ADC

∠DAC = 70° (given)

∠ADC = 30° (Proved above)

∠ACD = ∠ ACB+ ∠ BCD

∠ACD = 50° + 30°

∠ACD = 80°

∠ADC < ∠DAC < ∠ACD

AC < DC < AD (Side opposite to greater angle is longer and smaller angle is smaller)

AD > CD and AD > AC

Hence proved

**Question 4: Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?**

**Solution:**

We know that, Sum of 2 sides > third side

Case 1

2 + 3 > 7

2 + 3 ≯ 7

Case 2

2 + 7 > 3

9 > 3

Case 3

3 + 7 > 2

10 > 2

All the 3 case should be true.

Hence, it is not possible to form a triangle.

**Question 5: O is any point in the interior of ΔABC. Prove that**

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

**(ii) AB + BC + CA > OA + QB + OC**

**(iii) OA + OB + OC > (1/2) (AB + BC +CA)**

**Solution:**

(i) It is given that O is any point in the interior of ΔABC.

We know that, Sum of 2 sides > third side

In Δ ABC

AB + BC > AC

BC + AC > AB

AC + AB > BC

In ΔOBC

OB + OC > BC _______(i)

In ΔOAC

OA + OC > AC _______(ii)

In ΔOAB

OA + OB > AB _______(iii)

In ΔABD, we have

AB + AD > BD

AB + AD > BO + OD _______(iv) (BD = BO + OD)

In ΔODC, we have

OD + DC > OC _______ (v)

(i) By adding equation (iv) and (v)

AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC

we have

BC + BA > OA + OC ... (vii)

CA+ CB > OA + OB ... (viii)

(ii) Adding equation (vi), (vii) and (viii),

AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB

2AB + 2BC + 2CA > 2OA + 2OB + 2OC

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

AB + BC + CA > OA + OB + OC

(iii) Adding equations (i), (ii) and (iii)

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

2OA + 2OB + 2OC > AB + BC + CA

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

(OA + OB + OC) > (1/2)(AB + BC +CA)

**Question 6: Prove that the perimeter of a triangle is greater than the sum of its altitudes.**

**Solution:**

It is given that, ΔABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.

Therefore

AD ⊥ BC

AB > AD and AC > AD

AB + AC > 2AD _______(i) (perpendicular line segment is the shortest)

BE ⊥ AC

BA > BE and BC > BE

BA + BC > 2BE _______(ii) (perpendicular line segment is the shortest)

CF ⊥ AB

CA > CF and CB > CF

CA + CB > 2CF _______ (iii) (perpendicular line segment is the shortest)

By Adding equation (i), (ii) and (iii), we get

AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF

2AB + 2BC + 2CA > 2(AD + BE + CF)

AB + BC + CA > AD + BE + CF

Hence we can say that the sum of its altitudes are smaller then the perimeter of the triangle.

**Question 7: In Fig., prove that:**

**(i) CD + DA + AB + BC > 2AC**

**(ii) CD + DA + AB > BC**

**Solution:**

To prove

(i) CD + DA + AB + BC > 2AC

(ii) CD + DA+ AB > BC

From the given figure,

We know that, in a triangle sum of any two sides is greater than the third side

(i) In ΔABC, we have

AB + BC > AC ________(i)

In ΔADC, we have

CD + DA > AC ________ (ii)

By adding (i) and (ii), we get

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(ii) In Δ ABC, we have,

AB + AC > BC ________ (iii)

And in ΔADC, we have

CD + DA > AC

Add AB on both sides

CD + DA + AB > AC + AB

From equation (iii) and (iv), we get,

CD + DA + AB > AC + AB > BC

CD + DA + AB > BC

Hence proved

**Question 8: Which of the following statements are true (T) and which are false (F)?**

**(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.**

**(ii) Sum of any two sides of a triangle is greater than twice the median drown to the third side**

**(iii) Sum of any two sides of a triangle is greater than the third side.**

**(iv) Difference of any two sides of a triangle is equal to the third side.**

**(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it**

**(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.**

**Solution:**

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

**Question 9: Fill in the blanks to make the following statements true.**

**(i) In a right triangle the hypotenuse is the ___ side.**

**(ii) The sum of three altitudes of a triangle is ___ than its perimeter.**

**(iii) The sum of any two sides of a triangle is ___ than the third side.**

**(iv) If two angles of a triangle are unequal, then the smaller angle has the ___ side opposite to it.**

**(v) Difference of any two sides of a triangle is ___ than the third side.**

**(vi) If two sides of a triangle are unequal, then the larger side has ___ angle opposite to it.**

**Solution:**

(i) In a right triangle the hypotenuse is the largest side

(ii) The sum of three altitudes of a triangle is less than its perimeter.

(iii) The sum of any two sides of a triangle is greater than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.

(v) Difference of any two sides of a triangle is less than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.

**Exercise VSAQs**

**Question 1: In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.**

**Solution:**

It is given that, ∆ABC ≅ ∆DEF,

AB = DE and BC = EF

∠A = ∠D

∠B = ∠E

∠C = ∠F

**Question 2: In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠ E and ∠ C = ∠F. Are the two triangles necessarily congruent?**

**Solution:**

No.

Two triangles are not necessarily congruent, because we know only angle-angle-angle (AAA) criterion. This criterion can produce similar but not congruent triangles.

**Question 3: If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, C = 75°, DE = 2.5 cm, DF = 5 cm and D = 75°. Are two triangles congruent?**

**Solution:**

Yes

It is given that,

AC = DE (each 2.5cm)

BC = DF (each 5cm

∠D = ∠C (each 75°)

∆ABC ≅ ∆EDF (By SAS theorem triangle)

**Question 4: In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?**

**Solution:**

Yes

It is given that,

AB = AD (given)

BC = CD (given)

AC = AC (common)

∆ABC ≅ ∆ADC (By SSS theorem triangle)

**Question 5: In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?**

**Solution:**

Yes.

AC = CE (Given)

BC = CD (Given)

∠B = ∠D (each 90°)

∆ABC ≅ ∆CDE (By SSA criteria)

**Question 6: ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.**

**Solution:**

It is given that AB = AC and BE and CF are two medians.

In △CFB and △BEC

CE = BF (Since, AC = AB = (AC/2) = (AB/2) = CE = BF)

BC = BC (Common)

∠ECB = ∠FBC (Angle opposite to equal sides are equal)

△CFB ≅ △BEC (By SAS theorem)

BE = CF (CPCT)