Maths RD Sharma Chapter 9 Triangle and its Angles

Exercise 9.1 Page No: 9

Question 1: In a ΔABC, if ∠A = 550, ∠B = 400, find ∠C.
Solution:
Given: ∠A = 550, ∠B = 400
We know, sum of all angles of a triangle is 180 0
∠A + ∠B + ∠C = 1800
550 + 400 + ∠C=1800
950 + ∠C = 1800
∠C = 1800 − 95033
∠C = 850
Question 2: If the angles of a triangle are in the ratio 1:2:3, determine
three angles.
Solution:
Angles of a triangle are in the ratio 1:2:3 (Given)
Let the angles be x, 2x, 3x
Sum of all angles of triangles = 1800
x + 2x + 3x = 1800
6x = 1800
x = 1800/6
x = 300
Answer:
x = 300
2x = 2(30)0 = 600
3x = 3(30) 0 = 900
Question 3: The angles of a triangle are (x − 40) 0, (x − 20) 0 and (1/2 x −
10) 0. Find the value of x.
Solution:
The angles of a triangle are (x − 40)0, (x − 20) 0 and (1/2 x − 10) 0
Sum of all angles of triangle = 1800
(x − 40)0 + (x − 20) 0 + (1/2 x − 10) 0 = 1800
5/2 x – 700 = 180
5/2 x = 1800 + 700
5x = 2(250) 0
x = 5000/5
x = 1000
Question 4: The angles of a triangle are arranged in ascending order
of magnitude. If the difference between two consecutive angles is 10 0,
find the three angles.
Solution:
The difference between two consecutive angles is 10 0 (given)
Let x, x + 100, x + 200 be the consecutive angles
x + x + 100 + x + 200 = 1800
3x + 300 = 1800
3x = 1800– 300
3x = 1500
or x = 500
Again,
x + 100 = 500 + 100 = 600
x+200 = 500 + 200 = 700
Answer: Three angles are 500,600 and 700.
Question 5: Two angles of a triangle are equal and the third angle is
greater than each of those angles by 300. Determine all the angles of
the triangle.
Solution:
Two angles of a triangle are equal and the third angle is greater than each
of those angles by 300. (Given)
Let x, x, x + 300 be the angles of a triangle.
Sum of all angles in a triangle = 1800
x + x + x + 300 = 1800
3x + 300 = 1800
3x = 1500
or x = 500
And x + 300 = 500 + 300 = 800
Answer: Three angles are 500, 500 and 800.
Question 6: If one angle of a triangle is equal to the sum of the other
two, show that the triangle is a right angle triangle.
Solution:
One angle of a triangle is equal to the sum of the other two angles (given)
To Prove: One of the angles is 900
Let x, y and z are three angles of a triangle, where
z = x + y …(1)
Sum of all angles of a triangle = 1800
x + y + z = 1800
z + z = 1800 (Using equation (1))
2z = 1800
z = 900 (Proved)
Therefore, triangle is a right angled triangle.
Exercise 9.2 Page No: 9.18
Question 1: The exterior angles, obtained on producing the base of a
triangle both ways are 1040 and 1360. Find all the angles of the
triangle.
∠ACD = ∠ABC + ∠BAC [Exterior angle property]
Find ∠ABC:
∠ABC + ∠ABE = 1800 [Linear pair]
∠ABC + 1360 = 1800
∠ABC = 440
Find ∠ACB:
∠ACB + ∠ACD = 1800 [Linear pair]
∠ACB + 1040 = 1800
∠ACB = 760
Sum of all angles of a triangle = 1800
∠A + 440 + 760 = 1800
∠A = 1800 − 440 −760
∠ A = 600
Answer: Angles of a triangle are ∠ A = 600, ∠B = 440 and ∠C = 760
Question 2: In a △ABC, the internal bisectors of ∠B and ∠C meet at P
and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC +
∠BQC = 1800.
Solution:
In triangle ABC,
BP and CP are internal bisector of ∠B and ∠C respectively
=> External ∠B = 180o – ∠B
BQ and CQ are external bisector of ∠B and ∠C respectively.
=> External ∠C = 180 o – ∠C
In triangle BPC,
∠BPC + 1/2∠B + 1/2∠C = 180o
∠BPC = 180 o – (∠B + ∠C) …. (1)
In triangle BQC,
∠BQC + 1/2(180 o – ∠B) + 1/2(180 o – ∠C) = 180 o
∠BQC + 180 o – (∠B + ∠C) = 180 o
∠BPC + ∠BQC = 180 o [Using (1)]
Hence Proved.
Question 3: In figure, the sides BC, CA and AB of a △ABC have been
produced to D, E and F respectively. If ∠ACD = 1050 and ∠EAF = 450,
find all the angles of the △ABC.

∠BAC = ∠EAF = 450 [Vertically opposite angles]

∠ACD = 1800 – 1050 = 750 [Linear pair]
∠ABC = 1050 – 450 = 600 [Exterior angle property]
Question 4: Compute the value of x in each of the following figures:
(i)

Solution:
∠BAC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1120 = 680 [Linear pair]
Sum of all angles of a triangle = 1800
x = 1800 − ∠BAC − ∠ACB
= 1800 − 600 − 680 = 520
Answer: x = 520
(ii)

Solution:
∠ABC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1100 = 700 [Linear pair]
Sum of all angles of a triangle = 1800
x = ∠BAC = 1800 − ∠ABC − ∠ACB
= 1800 – 600 – 700 = 500
Answer: x = 500
(iii)

Solution:
∠BAE = ∠EDC = 520 [Alternate angles]
Sum of all angles of a triangle = 1800
x = 1800 – 400 – 520 = 1800 − 920 = 880
Answer: x = 880
(iv)

Solution:
CD is produced to meet AB at E.

∠BEC = 1800 – 450 – 500 = 850 [Sum of all angles of a triangle = 1800]

∠AEC = 1800 – 850 = 950 [Linear Pair]
Now, x = 950 + 350 = 1300 [Exterior angle Property]
Answer: x = 1300
Question 5: In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB.
Determine the value of x.
Solution:
Let ∠BAD = y, ∠BAC = 3y
∠BDA = ∠BAD = y (As AB=DB)
Now,
∠BAD + ∠BAC + 1080 = 1800 [Linear Pair]
y + 3y + 1080 = 1800
4y = 720
or y = 180
Now, In ΔADC
∠ADC + ∠ACD = 1080 [Exterior Angle Property]
x + 180 = 1800
x = 900
Answer: x = 900
Exercise VSAQs Page No: 9.21
Question 1: Define a triangle.
Solution: Triangle is a three-sided polygon that consists of three edges
and three vertices. The most important property of a triangle is that the sum
of the internal angles of a triangle is equal to 180 degrees.
Question 2: Write the sum of the angles of an obtuse triangle.
Solution: The sum of angles of obtuse triangle = 180°.
Question 3: In △ABC, if ∠B = 600, ∠C = 800 and the bisectors of angles
∠ABC and ∠ACB meet at point O, then find the measure of ∠BOC.
Solution:
∠B = 600, ∠C = 800 (given)
As per question:
∠OBC = 600/2 = 300 and
∠OCB = 800/2 = 400
In triangle BOC,
∠OBC + ∠OCB + ∠BOC = 180 0
[Sum of angles of a triangle = 1800]
300 + 400 + ∠BOC = 1800
∠BOC = 1100
Question 4: If the angles of a triangle are in the ratio 2:1:3, then find
the measure of smallest angle.
Solution:
Let angles of a triangles are 2x, x and 3x, where x is the smallest angle.
To find: measure of x.
As, Sum of angles of a triangle = 1800
2x + x + 3x = 1800
6x = 1800
x = 300. Answer
Question 5: If the angles A, B and C of △ABC satisfy the relation B – A
= C – B, then find the measure of ∠B.
Solution:
Sum of angles of a triangle = 1800
A + B + C = 1800 …(1)
B – A = C – B …(Given)
2B = C + A …(2)
(1) => 2B + B = 1800
3B =1800
Or B = 600

RD Sharma Chapter 10 Congruent Triangles

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:
Sides BA and CA have been produced such that BA = AD and CA = AE.
To prove: DE ∥ BC
Consider △ BAC and △DAE,
BA = AD and CA= AE (Given)
∠BAC = ∠DAE (vertically opposite angles)
By SAS congruence criterion, we have
△ BAC ≃ △ DAE
We know, corresponding parts of congruent triangles are equal
So, BC = DE and ∠DEA = ∠BCA, ∠EDA = ∠CBA
Now, DE and BC are two lines intersected by a transversal DB s.t.
∠DEA=∠BCA (alternate angles are equal)
Therefore, DE ∥ BC. Proved.
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,
As two sides of the triangle are equal, so △ PQR is an isosceles triangle
PQ = QR and ∠QPR = ∠QRP ……. (i)
Also, L and M are midpoints of PQ and QR respectively
PL = LQ = QM = MR = QR/2
Now, consider Δ LPN and Δ MRN, LP = MR
∠LPN = ∠MRN [From (i)]
∠QPR = ∠LPN and ∠QRP = ∠MRN
PN = NR [N is midpoint of PR]
By SAS congruence criterion,
Δ LPN ≃ Δ MRN
We know, corresponding parts of congruent triangles are equal.
So LN = MN
Proved.
Question 3: In figure, PQRS is a square and SRT is an equilateral
triangle. Prove that
(i) PT = QT (ii) ∠ TQR = 150
Solution:
Given: PQRS is a square and SRT is an equilateral triangle.
To prove:
(i) PT =QT and (ii) ∠ TQR =15°
Now,
PQRS is a square:
PQ = QR = RS = SP …… (i)
And ∠ SPQ = ∠ PQR = ∠ QRS = ∠ RSP = 90o
Also, △ SRT is an equilateral triangle:
SR = RT = TS …….(ii)
And ∠ TSR = ∠ SRT = ∠ RTS = 60°
From (i) and (ii)
PQ = QR = SP = SR = RT = TS ……(iii)
From figure,
∠TSP = ∠TSR + ∠ RSP = 60° + 90° = 150° and
∠TRQ = ∠TRS + ∠ SRQ = 60° + 90° = 150°
=> ∠ TSR = ∠ TRQ = 1500 ………………… (iv)
By SAS congruence criterion, Δ TSP ≃ Δ TRQ
We know, corresponding parts of congruent triangles are equal
So, PT = QT
Proved part (i).
Now, consider Δ TQR.
QR = TR [From (iii)]
Δ TQR is an isosceles triangle.
∠ QTR = ∠ TQR [angles opposite to equal sides]
Sum of angles in a triangle = 180∘
=> ∠QTR + ∠ TQR + ∠TRQ = 180°
=> 2 ∠ TQR + 150° = 180° [From (iv)]
=> 2 ∠ TQR = 30°
=> ∠ TQR = 150
Hence proved part (ii).
Question 4: Prove that the medians of an equilateral triangle are
equal.
Solution:
Consider an equilateral △ABC, and Let D, E, F are midpoints of BC, CA
and AB.
Here, AD, BE and CF are medians of △ABC.
Now,
D is midpoint of BC => BD = DC
Similarly, CE = EA and AF = FB
Since ΔABC is an equilateral triangle
AB = BC = CA …..(i)
BD = DC = CE = EA = AF = FB …………(ii)
And also, ∠ ABC = ∠ BCA = ∠ CAB = 60° ……….(iii)
Consider Δ ABD and Δ BCE
AB = BC [From (i)]
BD = CE [From (ii)]
∠ ABD = ∠ BCE [From (iii)]
By SAS congruence criterion,
Δ ABD ≃ Δ BCE
=> AD = BE ……..(iv)
[Corresponding parts of congruent triangles are equal in measure]
Now, consider Δ BCE and Δ CAF,
BC = CA [From (i)]
∠ BCE = ∠ CAF [From (ii)]
CE = AF [From (ii)]
By SAS congruence criterion,
Δ BCE ≃ Δ CAF
=> BE = CF …………..(v)
[Corresponding parts of congruent triangles are equal]
From (iv) and (v), we have
AD = BE = CF
Median AD = Median BE = Median CF
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:

To find: ∠ B and ∠ C.
Here, Δ ABC is an isosceles triangle since AB = AC
∠ B = ∠ C ……… (i)
[Angles opposite to equal sides are equal]
We know, sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
∠ A + ∠ B + ∠ B= 180° (using (i)
1200 + 2∠B = 1800
2∠B = 1800 – 1200 = 600
∠ B = 30o
Therefore, ∠ B = ∠ C = 30∘
Question 6: In a Δ ABC, if AB = AC and ∠ B = 70°, find ∠ A.
Solution:
Given: In a Δ ABC, AB = AC and ∠B = 70°
∠ B = ∠ C [Angles opposite to equal sides are equal]
Therefore, ∠ B = ∠ C = 70∘
Sum of angles in a triangle = 180∘
∠ A + ∠ B + ∠ C = 180o
∠ A + 70o + 70o = 180o
∠ A = 180o – 140o
∠ A = 40o

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:
In the figure,
RT = TS ……(i)
∠ 1 = 2 ∠ 2 ……(ii)
And ∠ 4 = 2 ∠ 3 ……(iii)
To prove: ΔRBT ≅ ΔSAT
Let the point of intersection RB and SA be denoted by O
∠ AOR = ∠ BOS [Vertically opposite angles]
or ∠ 1 = ∠ 4
2 ∠ 2 = 2 ∠ 3 [From (ii) and (iii)]
or ∠ 2 = ∠ 3 ……(iv)
Now in Δ TRS, we have RT = TS
=> Δ TRS is an isosceles triangle
∠ TRS = ∠ TSR ……(v)
But, ∠ TRS = ∠ TRB + ∠ 2 ……(vi)
∠ TSR = ∠ TSA + ∠ 3 ……(vii)
Putting (vi) and (vii) in (v) we get
∠ TRB + ∠ 2 = ∠ TSA + ∠ B
=> ∠ TRB = ∠ TSA [From (iv)]
Consider Δ RBT and Δ SAT
RT = ST [From (i)]
∠ TRB = ∠ TSA [From (iv)]
By ASA criterion of congruence, we have
Δ RBT ≅ Δ SAT
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: Lines AB and CD Intersect at O

Such that BC ∥ AD and

BC = AD …….(i)
To prove : AB and CD bisect at O.
First we have to prove that Δ AOD ≅ Δ BOC
∠OCB =∠ODA [AD||BC and CD is transversal]
AD = BC [from (i)]
∠OBC = ∠OAD [AD||BC and AB is transversal]
By ASA Criterion:
Δ AOD ≅ Δ BOC
OA = OB and OD = OC (By c.p.c.t.)
Therefore, AB and CD bisect each other at O.
Hence Proved.
Question 3: Solution:
Δ ABC is isosceles with AB = AC and BD and CE are bisectors of ∠ B and
∠ C We have to prove BD = CE. (Given)

Since AB = AC
=> ∠ABC = ∠ACB ……(i)
[Angles opposite to equal sides are equal]
Since BD and CE are bisectors of ∠ B and ∠ C
∠ ABD = ∠ DBC = ∠ BCE = ECA = ∠B/2 = ∠C/2 …(ii)
Now, Consider Δ EBC = Δ DCB
∠ EBC = ∠ DCB [From (i)]
BC = BC [Common side]
∠ BCE = ∠ CBD [From (ii)]
By ASA congruence criterion, Δ EBC ≅ Δ DCB
Since corresponding parts of congruent triangles are equal.
=> CE = BD
or, BD = CE
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:
In two right triangles one side and acute angle of one are equal to the
corresponding side and angles of the other. (Given)
To prove: Both the triangles are congruent.
Consider two right triangles such that
∠ B = ∠ E = 90o …….(i)
AB = DE …….(ii)
∠ C = ∠ F ……(iii)
Here we have two right triangles, △ ABC and △ DEF
From (i), (ii) and (iii),
By AAS congruence criterion, we have Δ ABC ≅ Δ DEF
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 [AD is a bisector of ∠ EAC]
∠1 = ∠3 [Corresponding angles]
and ∠2 = ∠4 [alternative angle]
From above, we have ∠3 = ∠4
This implies, AB = AC
Two sides AB and AC are equal.
=> Δ ABC 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:
Let Δ ABC be isosceles where AB = AC and ∠ B = ∠ C
Given: Vertex angle A is twice the sum of the base angles B and C. i.e., ∠
A = 2(∠ B + ∠ C)
∠ A = 2(∠ B + ∠ B)
∠ A = 2(2 ∠ B)
∠ A = 4(∠ B)
Now, We know that sum of angles in a triangle =180°
∠ A + ∠ B + ∠ C =180°
4 ∠ B + ∠ B + ∠ B = 180°
6 ∠ B =180°
∠ B = 30°
Since, ∠ B = ∠ C
∠ B = ∠ C = 30°
And ∠ A = 4 ∠ B
∠ A = 4 x 30° = 120°
Therefore, angles of the given triangle 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: Given that PQR is a triangle such that PQ = PR and S is any
point on the side PQ and ST ∥ QR.
To prove: PS = PT
Since, PQ= PR, so △PQR is an isosceles triangle.
∠ PQR = ∠ PRQ
Now, ∠ PST = ∠ PQR and ∠ PTS = ∠ PRQ
[Corresponding angles as ST parallel to QR]
Since, ∠ PQR = ∠ PRQ
∠ PST = ∠ PTS
In Δ PST,
∠ PST = ∠ PTS
Δ PST is an isosceles triangle.
Therefore, 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:
Consider Δ ADC and Δ CBA.
AB = CD [Given]
BC = AD [Given]
And AC = AC [Common side]
So, by SSS congruence criterion, we have
ΔADC≅ΔCBA
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:
Given: In Δ PQR, PQ = QR and L, M and N are the mid-points of the sides
PQ, QR and RP respectively
To prove: LN = MN
Join L and M, M and N, N and L
We have PL = LQ, QM = MR and RN = NP
[Since, L, M and N are mid-points of PQ, QR and RP respectively]
And also PQ = QR
PL = LQ = QM = MR = PN = LR …….(i)
[ Using mid-point theorem]
MN ∥ PQ and MN = PQ/2
MN = PL = LQ ……(ii)
Similarly, we have
LN ∥ QR and LN = (1/2)QR
LN = QM = MR ……(iii)
From equation (i), (ii) and (iii), we have
PL = LQ = QM = MR = MN = LN
This implies, 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:
Given: D is the midpoint of BC and PD = DQ in a triangle ABC.
To prove: ABC is isosceles triangle.

In △BDP and △CDQ

PD = QD (Given)
BD = DC (D is mid-point)
∠BPD = ∠CQD = 90o
By RHS Criterion: △BDP ≅ △CDQ
BP = CQ … (i) (By CPCT)
In △APD and △AQD
PD = QD (given)
AD = AD (common)
APD = AQD = 90 o
By RHS Criterion: △APD ≅ △AQD
So, PA = QA … (ii) (By CPCT)
Adding (i) and (ii)
BP + PA = CQ + QA
AB = AC
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 AS respectively s.t. BE = CF.
To prove: Δ ABC is isosceles

In Δ BCF and Δ CBE,

∠ BFC = CEB = 90o [Given]
BC = CB [Common side]
And CF = BE [Given]
By RHS congruence criterion: ΔBFC ≅ ΔCEB
So, ∠ FBC = ∠ EBC [By CPCT]
=>∠ ABC = ∠ ACB
AC = AB [Opposite sides to equal angles are equal in a triangle]
Two sides of triangle ABC are equal.
Therefore, Δ ABC is isosceles. Hence Proved.
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:
Consider an angle ABC and BP be one of the arm within the angle.
Draw perpendiculars PN and PM on the arms BC and BA.
In Δ BPM and Δ BPN,
∠ BMP = ∠ BNP = 90° [given]
BP = BP [Common side]
MP = NP [given]
By RHS congruence criterion: ΔBPM≅ΔBPN
So, ∠ MBP = ∠ NBP [ By CPCT]
BP is the angular bisector of ∠ABC.
Hence proved

Exercise 10.6 Page No: 10.66

Question 1: In Δ ABC, if ∠ A = 40° and ∠ B = 60°. Determine the
longest and shortest sides of the triangle.
Solution: In Δ ABC, ∠ A = 40° and ∠ B = 60°
We know, sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
40° + 60° + ∠ C = 180°
∠ C = 180° – 100° = 80°
∠ C = 80°
Now, 40° < 60° < 80°
=> ∠ A < ∠ B < ∠ C
=> ∠ C is greater angle and ∠ A is smaller angle.
Now, ∠ A < ∠ B < ∠ C
We know, side opposite to greater angle is larger and side opposite to
smaller angle is smaller.
Therefore, BC < AC < AB
AB is longest and BC is shortest side.
Question 2: In a Δ ABC, if ∠ B = ∠ C = 45°, which is the longest side?
Solution: In Δ ABC, ∠ B = ∠ C = 45°
Sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
∠ A + 45° + 45° = 180°
∠ A = 180° – (45° + 45°) = 180° – 90° = 90°
∠ A = 90°
=> ∠ B = ∠ C < ∠ A
Therefore, 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: In Δ ABC, side AB is produced to D so that BD = BC.
∠ B = 60°, and ∠ A = 70°

To prove: (i) AD > CD (ii) AD > AC

Construction: Join C and D
We know, sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
70° + 60° + ∠ C = 180°
∠ C = 180° – (130°) = 50°
∠ C = 50°
∠ ACB = 50° ……(i)
And also in Δ BDC
∠ DBC =180° – ∠ ABC = 180 – 60° = 120°
[∠DBA is a straight line]
and BD = BC [given]
∠ BCD = ∠ BDC [Angles opposite to equal sides are equal]
Sum of angles in a triangle =180°
∠ DBC + ∠ BCD + ∠ BDC = 180°
120° + ∠ BCD + ∠ BCD = 180°
120° + 2∠ BCD = 180°
2∠ BCD = 180° – 120° = 60°
∠ BCD = 30°
∠ BCD = ∠ BDC = 30° ….(ii)
Now, consider Δ ADC.
∠ DAC = 70° [given]
∠ ADC = 30° [From (ii)]
∠ ACD = ∠ ACB+ ∠ BCD = 50° + 30° = 80° [From (i) and (ii)]
Now, ∠ 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:
Lengths of sides are 2 cm, 3 cm and 7 cm.
A triangle can be drawn only when the sum of any two sides is greater than
the third side.
So, let’s check the rule.
2 + 3 ≯ 7 or 2 + 3 < 7
2+7>3
and 3 + 7 > 2
Here 2 + 3 ≯ 7
So, the triangle does not exit.

Exercise VSAQs
Question 1: In two congruent triangles ABC and DEF, if AB = DE and
BC = EF. Name the pairs of equal angles.
Solution:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF, then
∠A = ∠D, ∠B = ∠ E and ∠ 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.
Reason: 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 = 75o, DE = 2.5 cm, DF = 5 cm and D = 75o. Are two
triangles congruent?
Solution: Yes.
Reason: Given triangles are congruent as AC = DE = 2.5 cm, BC = DF = 5
cm and
∠D = ∠C = 75o.
By SAS theorem triangle ABC is congruent to triangle EDF.
Question 4: In two triangles ABC and ADC, if AB = AD and BC = CD.
Are they congruent?
Solution: Yes.
Reason: Given triangles are congruent as
AB = AD
BC = CD and
AC [ common side]
By SSS theorem triangle ABC is congruent to triangle ADC.
Question 5: In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60o,
∠C = 30 o and ∠D = 90 o. Are two triangles congruent?
Solution: Yes.
Reason: Given triangles are congruent
Here AC = CE
BC = CD
∠B = ∠D = 90°
By SSA criteria triangle ABC is congruent to triangle CDE.
Question 6: ABC is an isosceles triangle in which AB = AC. BE and CF
are its two medians. Show that BE = CF.
Solution: ABC is an isosceles triangle (given)
AB = AC (given)
BE and CF are two medians (given)
To prove: BE = CF
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)
By SAS theorem: △CFB ≅ △BEC
So, BE = CF (By c.p.c.t)