ICSE Class 9 Maths Selina Solutions Chapter 10:
Here are ICSE Class 9 Maths Selina Solutions Chapter 10. A student's time in class nine is crucial. Understanding the material covered in Class 9 is essential since Class 10 builds on it.
It is recommended that you complete the exercises in every chapter of the Selina publishing book to achieve high scores on the mathematics exam for Class 9. The Selina answers for Maths Class 9 aid students in better comprehending all of the material.
ICSE Class 9 Maths Selina Solutions Chapter 10 Overview
ICSE Class 9 Maths Selina Solutions Chapter 10 introduces isosceles triangles. In this ICSE Class 9 Maths Selina Solutions Chapter 10, students learn that an isosceles triangle has at least two sides of equal length, and the angles opposite these equal sides are also equal.
The ICSE Class 9 Maths Selina Solutions Chapter 10 explains important properties and theorems related to isosceles triangles, such as how the angles opposite the equal sides are equal and how this can be used to find missing angles and sides. By understanding these properties and practicing related problems, students gain the skills to solve various geometric problems involving isosceles triangles.
ICSE Class 9 Maths Selina Solutions Chapter 10 PDF
Below we have provided ICSE Class 9 Maths Selina Solutions Chapter 10 in detail. This chapter will help you to clear all your doubts regarding the chapter Inequalities. Students are advised to prepare from these ICSE Class 9 Maths Selina Solutions Chapter 10 before the examinations to perform better.
ICSE Class 9 Maths Selina Solutions Chapter 10 PDF
ICSE Class 9 Maths Selina Solutions Chapter 10 Isosceles Triangles
Below we have provided ICSE Class 9 Maths Selina Solutions Chapter 10 -
1. In the triangle
AB = AC
∠A = 48
o
and
∠ACD = 18
o
.
Show that BC = CD.
Solution:
In ∆ABC, we have
∠
BAC +
∠
ACB +
∠
ABC = 180
0
48
0
+
∠
ACB +
∠
ABC = 180
0
But,
∠
ACB =
∠
ABC [Given, AB = AC]
2
∠
ABC = 180
0
– 48
0
2
∠
ABC = 132
0
∠
ABC = 66
0
=
∠
ACB ……(i)
∠
ACB = 66
0
∠
ACD +
∠
DCB = 66
0
18
0
+
∠
DCB = 66
0
∠
DCB = 48
0
………(ii)
Now, In ∆DCB,
∠
DBC = 66
0
[From (i), Since
∠
ABC =
∠
DBC]
∠
DCB = 48
0
[From (ii)]
∠
BDC = 180
0
– 48
0
– 66
0
∠
BDC = 66
0
Since
∠
BDC =
∠
DBC
Therefore, BC = CD
Equal angles have equal sides opposite to them.
2. Calculate:
(i) ∠ADC
(ii) ∠ABC
(iii) ∠BAC
Solution:
Given: ∠ACE = 130
0
; AD = BD = CD
Proof:
(i) ∠ACD + ∠ACE = 180
o
[DCE is a straight line]
∠ACD = 180
o
– 130
o
∠ACD = 50
o
Now,
CD = AD
∠ACD = ∠DAC = 50
o
… (i) [Since angles opposite to equal sides are equal]
In ∆ADC,
∠ACD = ∠DAC = 50
o
∠ACD + ∠DAC + ∠ADC = 180
o
50
o
+ 50
o
+ ∠ADC = 180
o
∠ADC = 180
o
– 100
o
∠ADC = 80
o
(ii) ∠ADC = ∠ABD + ∠DAB [Exterior angle is equal to sum of opposite interior angles]
But, AD = BD
∴ ∠DAB=∠ABD
80
o
= ∠ABD + ∠ABD
2∠BD = 80
O
∠ABD = 40
O
= ∠DAB … (ii)
(iii) We have,
∠BAC = ∠DAB + ∠DAC
Substituting the values from (i) and (ii),
∠BAC = 40
O
+ 50
O
Hence, ∠BAC = 90
O
3. In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate:
(i) ∠CDE
(ii) ∠DCE
Solution:
Given, ∠FAB = 128
O
∠BAC + ∠FAB = 180
O
[As FAC is a straight line]
∠BAC = 180
O
– 128
O
∠BAC = 52
O
In
∆ABC, we have
∠A = 52
O
∠B = ∠C [Given AB = AC and angles opposite to equal sides are equal]
Now, by angle sum property
∠A + ∠B + ∠C =180
O
∠A + ∠B + ∠B = 180
O
52
O
+ 2∠B = 180
O
2∠B = 128
O
∠B = 64
O
= ∠C… (i)
∠B = ∠ADE [Given DE ll BC]
(i) Now, ∠ADE + ∠CDE + ∠B = 180
O
[As ADB is a straight line]
64
O
+ ∠CDE + 64
O
= 180
O
∠CDE = 180
O
+ 128
O
∠CDE = 52
O
(ii) Given DE ll BC and DC is the transversal
∠CDE = ∠DCB = 52
o
… (ii)
Also, ∠ECB = 64
o
… [From (i)]
But,
∠ECB = ∠DCE +∠DCB
64
o
= ∠DCE + 52
o
∠DCE = 64
o
– 52
o
∠DCE = 12
o
4. Calculate x:
Solution:
(i) Let the triangle be ABC and the altitude be AD.
In ∆ABD, we have
∠DBA = ∠DAB = 37
o
[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB [Exterior angle is equal to the sum of opposite interior angles]
∠CDA = 37
o
+ 37
o
∴ ∠CDA = 74
o
Now, in ∆ADC, we have
∠CDA = ∠CAD = 74
o
[Given CD = AC and angels opposite to equal sides are equal]
Now, by angle sum property
∠CAD + ∠CDA + ∠ACD = 180
o
74
o
+ 74
o
+ x = 180
o
x = 180
o
– 148
o
x = 32
o
(ii) Let triangle be ABC and altitude be AD.
In ∆ABD, we have
∠DBA = ∠DAB = 50
o
[Given BD = AD and angles opposite to equal sides are equal]
Now,
∠CDA = ∠DBA + ∠DAB [Exterior angle is equal to the sum of opposite interior angles]
∠CDA = 50
o
+ 50
o
∴ ∠CDA = 100
o
In ∆ADC, we have
∠DAC = ∠DCA = x [Given AD = DC and angels opposite to equal sides are equal]
So, by angle sum property
∠DAC + ∠DCA + ∠ADC = 180
o
x + x + 100
o
= 180
o
2x = 80
o
x = 40
o
5. In the figure, given below, AB = AC. Prove that: ∠BOC = ∠ACD.
Solution:
Let’s assume ∠ABO = ∠OBC = x and ∠ACO = ∠OCB = y
In ABC, we have
∠BAC = 180
o
– 2x – 2y…(i)
As, ∠B = ∠C [Since, AB = AC]
½ ∠B = ½ ∠C
⇒ x = y
Now,
∠ACD = 2x + ∠BAC [Exterior angle is equal to sum of opposite interior angle]
= 2x + 180
o
– 2x – 2y [From (i)]
∠ACD = 180
o
– 2y… (ii)
In ∆OBC, we have
∠BOC = 180
o
– x – y
∠BOC = 180
o
– y – y [Since x = y]
∠BOC = 180
o
– 2y… (iii)
Thus, from (ii) and (iii) we get
∠BOC = ∠ACD
6. In the figure given below, LM = LN; ∠PLN = 110
o
. Calculate:
(i) ∠LMN
(ii) ∠MLN
Solution:
Given, LM = LN and ∠PLN = 110
o
(i) We know that the sum of the measure of all the angles of a quadrilateral is 360
o
.
In quad. PQNL,
∠QPL + ∠PLN +LNQ + ∠NQP = 360
o
90
o
+ 110
o
+ ∠LNQ + 90
o
= 360
o
∠LNQ = 360
o
– 290
o
∠LNQ = 70
o
∠LNM = 70
o
… (i)
In ∆LMN, we have
LM = LN [Given]
⇒ ∠LNM = ∠LMN [ Angles opposite to equal sides are equal]
∠LMN = 70
o
…(ii) [From (i)]
(ii) In ∆LMN, we have
∠LMN + ∠LNM + ∠MLN = 180
o
But, ∠LNM = ∠LMN = 70
o
[From (i) and (ii)]
⇒ 70
o
+ 70
o
+ ∠MLN = 180
o
∠MLN = 180
o
– 140
o
∴ ∠MLN = 40
o
7. An isosceles triangle ABC has AC = BC. CD bisects AB at D and ∠CAB = 55
o
.
Find: (i) ∠DCB (ii) ∠CBD.
Solution:
In ∆ABC, we have
AC = BC [Given]
So, ∠CAB = ∠CBD [Angles opposite to equal sides are equal]
⇒ ∠CBD = 55
o
In ∆ABC, we have
∠CBA + ∠CAB + ∠ACB = 180
o
But, ∠CAB = ∠CBA = 55
o
55
o
+ 55
o
+ ∠ACB = 180
o
∠ACB = 180
o
– 110
o
∠ACB = 70
o
Now,
In ∆ACD and ∆BCD, we have
AC = BC [Given]
CD = CD [Common]
AD = BD [Given that CD bisects AB]
∴ ∆ACD ≅ ∆BCD
So, By CPCT
∠DCA = ∠DCB
∠DCB = ∠ACB/2 = 70
o
/2
Thus, ∠DCB = 35
o
8. Find x:
Solution:
Let’s put markings to the figure as following:
In ∆ABC, we have
AD = AC [Given]
∴ ∠ADC = ∠ACD [Angles opposite to equal sides are equal]
So, ∠ADC = 42
o
Now,
∠ADC = ∠DAB + DBA [Exterior angle is equal to the sum of opposite interior angles]
But,
∠DAB = ∠DBA [Given: BD = DA]
∴ ∠ADC = 2∠DBA
2∠DBA = 42
o
∠DBA = 21
o
To find x:
x = ∠CBA + ∠BCA [Exterior angle is equal to the sum of opposite interior angles]
We know that,
∠CBA = 21
o
∠BCA = 42
o
⇒ x = 21
o
+ 42
o
∴ x = 63
o
9. In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.
Solution:
In ∆ABC and ∆DBC, we have
BD = BD [Common]
∠BDA = ∠BDC [Each equal to 90
o
]
∠ABD = ∠DBC [BD bisects ∠ABC]
∴ ∆ABD ≅ ∆DBC [ASA criterion]
Therefore, by CPCT
AD = DC
x + 1 = y + 2
x = y + 1… (i)
And, AB = BC
3x + 1 = 5y – 2
Substituting the value of x from (i), we get
3(y+1) + 1 = 5y – 2
3y + 3 + 1 = 5y – 2
3y + 4 = 5y – 2
2y = 6
y = 3
Putting y = 3 in (i), we get
x = 3 + 1
∴ x = 4
10. In the given figure; AE // BD, AC // ED and AB = AC. Find ∠a, ∠b and ∠c.
Solution:
Let’s assume points P and Q as shown below:
Given, ∠PDQ = 58
o
∠PDQ = ∠EDC = 58
o
[Vertically opposite angles]
∠EDC = ∠ACB = 58
o
[Corresponding angles ∵ AC ll ED]
In ∆ABC, we have
AB = AC [Given]
∴ ∠ACB = ∠ABC = 58
o
[Angles opposite to equal sides are equal]
Now,
∠ACB + ∠ABC + ∠BAC = 180
o
58
o
+ 58
o
+ a = 180
o
∠a = 180
o
– 116
o
∠a = 64
o
Since, AE ll BD and AC is the transversal
∠ABC = ∠b [Corrosponding angles]
∴ ∠b = 58
o
Also, since AE ll BD and ED is the transversal
∠EDC = ∠c [Corrosponding angles]
∴ ∠c = 58
o
11. In the following figure; AC = CD, AD = BD and ∠C = 58
o
.
Find ∠CAB.
Solution:
In ∆ACD, we have
AC = CD [Given]
∴ ∠CAD = ∠CDA [Angles opposite to equal sides are equal]
And,
∠ACD = 58
o
[Given]
By angle sum property, we have
∠ACD + ∠CDA + ∠CAD = 180
o
58
o
+ 2∠CAD = 180
o
2∠CAD = 122
o
∠CAD = ∠CDA = 61
o
… (i)
Now,
∠CDA = ∠DAB + ∠DBA [Exterior angles is equal to sum of opposite interior angles]
But,
∠DAB = ∠DBA [Given, AD = DB]
So, ∠DAB + ∠DAB = ∠CDA
2∠DAB = 61
o
∠DAB = 30.5
o
… (ii)
In ∆ABC, we have
∠CAB = ∠CAD + ∠DAB
∠CAB = 61
o
+ 30.5
o
[From (i) and (ii)]
∴ ∠CAB = 91.5
o
12. In the figure of Q.11 is given above, if AC = AD = CD = BD; find angle ABC.
Solution:
In ∆ACD, we have
AC = AD = CD [Given]
Hence, ACD is an equilateral triangle
∴ ∠ACD = ∠CDA = ∠CAD = 60
o
Now,
∠CDA = ∠DAB + ∠ABD [Exterior angle is equal to sum of opposite interior angles]
But,
∠DAB = ∠ABD [Given, AD = DB]
So, ∠ABD + ∠ABD = ∠CDA
2∠ABD = 60
o
∴ ∠ABD = ∠ABC = 30
o
13. In ∆ABC; AB = AC and ∠A: ∠B = 8: 5; find ∠A.
Solution:
Let, ∠A = 8x and ∠B = 5x
Given, In ∆ABC
AB = AC
So, ∠B = ∠C = 5x [Angles opp. to equal sides are equal]
Now, by angle sum property
∠A + ∠B +C = 180
o
8x + 5x + 5x = 180
o
18x = 180
o
x = 10
o
Thus, as ∠A = 8x
∠A = 8 × 10
o
∴ ∠A = 80
o
14. In triangle ABC; ∠A = 60
o
, ∠C = 40
o
, and bisector of angle ABC meets side AC at point P. Show that BP = CP.
Solution:
In ∆ABC, we have
∠A = 60
o
∠C = 40
o
∴ ∠B = 180
o
– 60
o
– 40
o
[By angle sum property]
∠B = 80
o
Now, as BP is the bisector of ∠ABC
∴ ∠PBC = ∠ABC/2
∠PBC = 40
o
In ∆PBC, we have
∠PBC = ∠PCB = 40
o
∴ BP = CP [Sides opposite to equal angles are equal]
15. In triangle ABC; angle ABC = 90
o
and P is a point on AC such that ∠PBC = ∠PCB. Show that: PA = PB.
Solution:
Let’s assume ∠PBC = ∠PCB = x
In the right-angled triangle ABC,
∠ABC = 90
o
∠ACB = x
∠BAC = 180
o
– (90
o
+ x) [By angle sum property]
∠BAC = (90
o
– x) …(i)
And
∠ABP = ∠ABC – ∠PBC
∠ABP = 90
o
– x …(ii)
Thus, in the ∆ABP from (i) and (ii), we have
∠BAP = ∠ABP
Therefore, PA = PB [sides opp. to equal angles are equal]
16. ABC is an equilateral triangle. Its side BC is produced upto point E such that C is mid-point of BE. Calculate the measure of angles ACE and AEC.
Solution:
Given, ∆ABC is an equilateral triangle
So, AB = BC = AC
∠ABC = ∠CAB = ∠ACB = 60
o
Now, as sum of two non-adjacent interior angles of a triangle is equal to the exterior angle
∠CAB + ∠CBA = ∠ACE
60
o
+ 60
o
= ∠ACE
∠ACE = 120
o
Now,
∆ACE is an isosceles triangle with AC = CF
∠EAC = ∠AEC
By angle sum property, we have
∠EAC + ∠AEC + ∠ACE = 180
o
2∠AEC + 120
o
= 180
o
2∠AEC = 180
o
– 120
o
∠AEC = 30
o
17. In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.
Solution:
From given, we get
∆DBC is an isosceles triangle
⇒ CD = DB
∠DBC = ∠DCB [If two sides of a triangle are equal, then angles opposites to them are equal]
And, ∠B = ∠DBC = ∠DCB = 28
o
By angle sum property, we have
∠DCB + ∠DBC + ∠BCD = 180
o
28
o
+ 28
o
+ ∠BCD = 180
o
∠BCD = 180
o
– 56
o
∠BCD = 124
o
As sum of two non-adjacent interior angles of a triangle is equal to the exterior angle, we have
∠DBC + ∠DCB = ∠DAC
28
o
+ 28
o
= 56
o
∠DAC
= 56
o
Now,
∆ACD is an isosceles triangle with AC = DC
⇒ ∠ADC = ∠DAC = 56
o
∠ADC + ∠DAC +∠DCA = 180
o
[By angle sum property]
56
o
+ 56
o
+ ∠DCA = 180
o
∠DCA = 180
o
– 112
o
∠DCA = 64
o
Thus, ∠ACD = 64
o
18. In the given figure, AD = AB = AC, BD is parallel to CA and ∠ACB = 65°. Find ∠DAC.
Solution:
From figure, it’s seen that
∆ABC is an isosceles triangle with AB = AC
⇒ ∠ACB = ∠ABC
As ∠ACB = 65
o
[Given]
∴ ∠ABC = 65
o
By angle sum property, we have
∠ACB + ∠CAB + ∠ABC = 180
o
65
o
+ 65
o
+ ∠CAB = 180
o
∠CAB = 180
o
– 130
o
∠CAB = 50
o
As BD is parallel to CA, we have
∠CAB = ∠DBA as they are alternate angles
⇒ ∠CAB = ∠DBA = 50
o
Again, from figure, it’s seen that
∆ADB is an isosceles triangle with AD = AB.
⇒ ∠ADB = ∠DBA = 50
o
By angle sum property, we have
∠ADB + ∠DAB + ∠DBA = 180
o
50
o
+ ∠DAB + 50
o
= 180
o
∠DAB = 180
o
– 100
o
∠DAB = 80
o
Now,
∠DAC = ∠CAB + ∠DAB
∠DAC = 50
o
– 80
o
∠DAC = 130
o
19. Prove that a triangle ABC is isosceles, if:
(i) altitude AD bisects angles BAC, or
(ii) bisector of angle BAC is perpendicular to base BC.
Solution:
(i) In
ΔABC, if the altitude AD bisect ∠BAC.
Then, to prove: ΔABC is isosceles.
In
ΔADB and ΔADC, we have
∠BAD = ∠CAD (AD is bisector of ∠BAC)
AD = AD (Common)
∠ADB = ∠ADC (Each equal to 90°)
Therefore, ΔADB ≅ ΔADC by ASA congruence criterion
So, by CPCT
AB = AC
Hence, ΔABC is an isosceles.
(ii) In Δ ABC, the bisector of ∠BAC is perpendicular to the base BC.
Then, to prove: ΔABC is isosceles.
In ΔADB and ΔADC,
∠BAD = ∠CAD (AD is bisector of ∠BAC)
AD = AD (Common)
∠ADB = ∠ADC (Each equal to 90°)
Therefore, ΔADB ≅ ΔADC by ASA congruence criterion
Thus, by CPCT
AB = AC
Hence, ΔABC is an isosceles.
20. In the given figure; AB = BC and AD = EC.
Prove that: BD = BE.
Solution:
In ΔABC, we have
AB = BC (Given)
So, ∠BCA = ∠BAC (Angles opposite to equal sides are equal)
⇒ ∠BCD = ∠BAE ….(i)
Given, AD = EC
AD + DE = EC + DE (Adding DE on both sides)
⇒ AE = CD ….(ii)
Now, in ΔABE and ΔCBD, we have
AB = BC (Given)
∠BAE = ∠BCD [From (i)]
AE = CD [From (ii)]
Therefore, ΔABE ≅ ΔCBD by SAS congruence criterion
So, by CPCT
BE = BD
ICSE Class 9 Maths Selina Solutions Chapter 10 Exercise 10B
1. If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.
Solution:
Construction: AB is produced to D and AC is produced to E so that exterior angles ∠DBC and ∠ECB are formed.
In ∆ABC, we have
AB = AC [Given]
∴ ∠C = ∠B …(i) [Angles opposite to equal sides are equal]
Since, ∠B and ∠C are acute they cannot be right angles or obtuse angles
Now,
∠ABC + ∠DBC = 180
0
[ABD is a straight line]
∠DBC = 180
0
– ∠ABC
∠DBC = 180
0
– ∠B …(ii)
Similarly,
∠ACB + ECB = 180
0
[ABD is a straight line]
∠ECB = 180
0
– ∠ACB
∠ECB = 180
0
– ∠C …(iii)
∠ECB = 180
0
– ∠B …(iv) [from (i) and (iii)]
∠DBC = ∠ECB [from (ii) and (iv)]
Now,
∠DBC = 180
0
– ∠B
But, ∠B is an acute angle
⇒ ∠DBC = 180
0
– (acute angle) = obtuse angle
Similarly,
∠ECB = 180
0
– ∠C
But, ∠C is an acute angle
⇒ ∠ECB = 180
0
– (acute angle) = obtuse angle
Therefore, exterior angles formed are obtuse and equal.
2. In the given figure, AB = AC. Prove that:
(i) DP = DQ
(ii) AP = AQ
(iii) AD bisects ∠A
Solution:
Construction: Join AD.
In ∆ABC, we have
AB = AC [Given]
∴ ∠C = ∠B …(i) [ Angles opposite to equal sides are equal]
(i) In ∆BPD and ∆CQD, we have
∠BPD = ∠CQD [Each = 90
o
]
∠B = ∠C [Proved]
BD = DC [Given]
Thus, ∆BPD ≅ ∆CQD by AAS congruence criterion
∴ DP = DQ by CPCT
(ii) Since, ∆BPD ≅ ∆CQD
Therefore, BP = CQ [CPCT]
Now,
AB = AC [Given]
AB – BP = AC – CQ
AP = AQ
(iii) In ∆APD and ∆AQD, we have
DP = DQ [Proved]
AD = AD [Common]
AP = AQ [Proved]
Thus, ∆APD ≅ ∆AQD by SSS congruence criterion
∠PAD = ∠QAD by CPCT
Hence, AD bisects angle A.
3. In triangle ABC, AB = AC; BE
⊥
AC and CF
⊥
AB. Prove that:
(i) BE = CF
(ii) AF = AE
Solution:
(i) In ∆AEB and ∆AFC, we have
∠A = ∠A [Common]
∠AEB = ∠AFC = 90
o
[ Given : BE ⊥ AC and CE ⊥ AB]
AB = AC [Given]
Thus, ∆AEB ≅ ∆AFC by AAS congruence criterion
∴ BE = CF by CPCT
(ii) Since, ∆AEB ≅ ∆AFC
∠ABE = ∠AFC
∴ AF= AE by CPCT
4. In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD. Prove that: ∠BCD = 90
o
Solution:
Construction: Join CD.
In ∆ABC, we have
AB = AC [Given]
∴ ∠C = ∠B … (i) [ Angles opposite to equal sides are equal]
In ∆ACD, we have
AC = AD [Given]
∴ ∠ADC = ∠ACD … (ii)
Adding (i) and (ii), we get
∠B + ∠ADC = ∠C + ∠ACD
∠B + ∠ADC = ∠BCD … (iii)
In ∆BCD, we have
∠B + ∠ADC +∠BCD = 180
o
∠BCD + ∠BCD = 180
o
[From (iii)]
2∠BCD = 180
o
∴ ∠BCD = 90
o
5. (i) In ∆ABC, AB = AC and ∠A= 36°. If the internal bisector of ∠C meets AB at point D, prove that AD = BC.
(ii) If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.
Solution:
Given, AB = AC and ∠A = 36
o
So, ∆ABC is an isosceles triangle.
∠B = ∠C = (180
0
– 36
o
)/2 = 72
o
∠ACD = ∠BCD = 36
o
[∵ CD is the angle bisector of ∠C]
Now, ∆ADC is an isosceles triangle as ∠DAC = ∠DCA = 36
o
∴ AD = CD …(i)
In ∆DCB, by angle sum property we have
∠CDB = 180
o
– (∠DCB + ∠DBC)
= 180
o
– (36
o
+ 72
o
)
= 180
o
– 108
o
= 72
o
Now, ∆DCB is an isosceles triangle as ∠CDB = ∠CBD = 72
o
∴ DC = BC …(ii)
From (i) and (ii), we get
AD = BC
– Hence Proved.
6. Prove that the bisectors of the base angles of an isosceles triangle are equal.
Solution:
In ∆ABC, we have
AB = AC [Given]
∴ ∠C = ∠B …(i) [Angles opposite to equal sides are equal]
½∠C = ½∠B
⇒ ∠BCF = ∠CBE …(ii)
Now, in ∆BCE and ∆CBF, we have
∠C = ∠B [From (i)]
∠BCF = ∠CBE [From (ii)]
BC = BC [Common]
∴ ∆BCE ≅ ∆CBF by AAS congruence criterion
Thus, BE = CF by CPCT
7. In the given figure, AB = AC and ∠DBC = ∠ECB = 90
o
Prove that:
(i) BD = CE
(ii) AD = AE
Solution:
In ∆ABC, we have
AB = AC [Given]
∴ ∠ACB = ∠ABC [Angles opposite to equal sides are equal]
⇒ ∠ABC = ∠ACB … (i)
∠DBC = ∠ECB = 90
o
[Given]
⇒ ∠DBC = ∠ECB …(ii)
Subtracting (i) from (ii), we get
∠DCB – ∠ABC = ∠ECB – ∠ACB
∠DBA = ∠ECA … (iii)
Now,
In ΔDBA and ΔECA, we have
∠DBA = ∠ECA [From (iii)]
∠DAB = ∠EAC [Vertically opposite angles]
AB = AC [Given]
∴ ΔDBA ≅ ΔECA by ASA congruence criterion
Thus, by CPCT
BD = CE
And, also
AD = AE
8. ABC and DBC are two isosceles triangles on the same side of BC. Prove that:
(i) DA (or AD) produced bisects BC at right angle.
(ii) ∠BDA = ∠CDA.
Solution:
DA is produced to meet BC in L
In ∆ABC, we have
AB = AC [Given]
∴ ∠ACB = ∠ABC … (i) [Angles opposite to equal sides are equal]
In ∆DBC, we have
DB = DC [Given]
∴ ∠DCB = ∠DBC … (ii) [Angles opposite to equal sides are equal]
Subtracting (i) from (ii), we get
∠DCB – ∠ACB = ∠DBC – ∠ABC
∠DCA = ∠DBA …(iii)
Now,
In ∆DBA and ∆DCA, we have
DB = DC [Given]
∠DBA = ∠DCA [From (iii)]
AB = AC [Given]
∴ ∆DBA ≅ ∆DCA by SAS congruence criterion
∠BDA = ∠CDA …(iv) [By CPCT]
In ∆DBA, we have
∠BAL = ∠DBA + ∠BDA …(v) [Exterior angle = sum of opposite interior angles]
From (iii), (iv) and (v), we get
∠BAL = ∠DCA + ∠CDA …(vi) [Exterior angle = sum of opposite interior angles]
In ∆DCA, we have
∠CAL = ∠DCA + ∠CDA …(vi)
From (vi) and (vii)
∠BAL = ∠CAL …(viii)
In ∆BAL and ∆CAL,
∠BAL = ∠CAL [From (viii)]
∠ABL = ∠ACL [From (i)
AB = AC [Given]
∴ ∆BAL ≅ ∆CAL by ASA congruence criterion
So, by CPCT
∠ALB = ∠ALC
And, BL = LC …(ix)
Now,
∠ALB + ∠ALC = 180
o
∠ALB + ∠ALB = 180
o
[Using (ix)]
2∠ALB = 180
o
∠ALB = 90
o
∴ AL ⊥ BC
Or DL ⊥ BC and BL ⊥ LC
Therefore, DA produced bisects BC at right angle.
9. The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.
Solution:
In ∆ABC, we have AB = AC
∠B = ∠C [Angles opposite to equal sides are equal]
½∠B = ½∠C
∠OBC = ∠OCB …(i)
⇒ OB = OC …(ii) [Sides opposite to equal angles are equal]
Now,
In ∆ABO and ∆ACO, we have
AB = AC [Given]
∠OBC = ∠OCB [From (i)]
OB = OC [From (ii)]
Thus, ∆ABO ≅ ∆ACO by SAS congruence criterion
So, by CPCT
∠BAO = ∠CAO
Therefore, AO bisects ∠BAC.
10. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.
Solution:
In ∆ABC, we have
AB = AC [Given]
∠C = ∠B … (i) [Angles opposite to equal sides are equal]
Now,
½ AB = ½ AC
BF = CE … (ii)
In ∆BCE and ∆CBF, we have
∠C = ∠B [From (i)]
BF = CE [From (ii)]
BC = BC [Common]
∴ ∆BCE ≅ ∆CBF by SAS congruence criterion
So, CPCT
BE = CF
11. Use the given figure to prove that, AB = AC.
Solution:
In ∆APQ, we have
AP = AQ [Given]
∴ ∠APQ = ∠AQP …(i) [Angles opposite to equal sides are equal]
In ∆ABP, we have
∠APQ = ∠BAP + ∠ABP …(ii) [Exterior angle is equal to sum of opposite interior angles]
In ∆AQC, we have
∠AQP = ∠CAQ + ∠ACQ …(iii) [Exterior angle is equal to sum of opposite interior angles]
From (i), (ii) and (iii), we get
∠BAP + ∠ABP = ∠CAQ + ∠ACQ
But, ∠BAP = ∠CAQ [Given]
∠CAQ + ABP = ∠CAQ + ∠ACQ
∠ABP = ∠CAQ + ∠ACQ – ∠CAQ
∠ABP = ∠ACQ
∠B = ∠C
So, in ∆ABC, we have
∠B = ∠C
⇒ AB = AC [Sides opposite to equal angles are equal]
12. In the given figure; AE bisects exterior angle CAD and AE is parallel to BC.
Prove that: AB = AC.
Solution:
Since, AE || BC and DAB is the transversal
∴ ∠DAE = ∠ABC = ∠B [Corresponding angles]
Since, AE || BC and AC is the transversal
∠CAE = ∠ACB = ∠C [Alternate angles]
But, AE bisects ∠CAD
∴ ∠DAE = ∠CAE
∠B = ∠C
⇒ AB = AC [Sides opposite to equal angles are equal]
13. In an equilateral triangle ABC; points P, Q and R are taken on the sides AB, BC and CA respectively such that AP = BQ = CR. Prove that triangle PQR is equilateral.
Solution:'
Given, AB = BC = CA (Since, ABC is an equilateral triangle) …(i)
and AP = BQ = CR …(ii)
Subtracting (ii) from (i), we get
AB – AP = BC – BQ = CA – CR
BP = CQ = AR …(iii)
∴ ∠A = ∠B = ∠C …(iv) [Angles opposite to equal sides are equal]
In ∆BPQ and ∆CQR, we have
BP = CQ [From (iii)]
∠B = ∠C [From (iv)]
BQ = CR [Given]
∴ ∆BPQ ≅ ∆CQR by SAS congruence criterion
So, PQ = QR [by CPCT] … (v)
In ∆CQR and ∆APR, we have
CQ = AR [From (iii)]
∠C = ∠A [From (iv)]
CR = AP [Given]
∴ ∆CQR ≅ ∆APR by SAS congruence criterion
So, QR = PR [By CPCT] … (vi)
From (v) and (vi), we get
PQ = QR = PR
Therefore, PQR is an equilateral triangle.
Benefits of ICSE Class 9 Maths Selina Solutions Chapter 10
Studying Chapter 10 on Isosceles Triangles from the ICSE Class 9 Maths Selina Solutions offers several benefits:
Understanding Basic Properties
: Students gain a clear understanding of the fundamental properties of isosceles triangles, including the fact that the angles opposite the equal sides are equal. This foundational knowledge is crucial for solving more complex geometric problems.
Enhanced Problem-Solving Skills
: By applying the properties and theorems related to isosceles triangles, students develop strong problem-solving skills. They learn how to use these properties to find missing angles and sides in various geometric configurations.
Foundation for Advanced Topics
: Mastery of isosceles triangles provides a solid foundation for studying more advanced geometric concepts. Understanding these basic principles helps in grasping more complex theorems and proofs in later chapters.
Preparation for Exams
: The ICSE Class 9 Maths Selina Solutions Chapter 10 equips students with the knowledge and practice needed for exams. It covers essential theorems and their applications, which are often tested in ICSE exams.
Improved Analytical Thinking
: Working through problems involving isosceles triangles enhances analytical and logical thinking skills. Students learn to approach problems systematically and apply theoretical knowledge in practical scenarios.
