# Concise Selina Solutions for Class 9 Maths Chapter 10- Isosceles Triangle

Selina Solutions for Class 9 Maths Chapter 10 Isosceles Triangle are provided here. Class 9 is an important phase in a studentâ€™s life. It is crucial to understand the concepts taught in Class 9 as these concepts are continued in Class 10. To score good marks in Class 9 Mathematics examination, it is advised to solve questions provided in each exercise across all the chapters in the book by Selina publication. This Selina solutions for Class 9 Maths helps students in understanding all the concepts in a better way. Download pdf of Class 9 Maths Chapter 10 Selina Solutions from the given link.

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Exercise 10(A)

1. In the figure alongside,

AB = AC

âˆ A = 48oÂ and

âˆ ACD = 18o.

Show that BC = CD.

Solution:

In âˆ†ABC, we have

âˆ BAC +Â âˆ ACB +Â âˆ ABC = 1800

480Â +Â âˆ ACB +Â âˆ ABC = 1800

But,Â âˆ ACB =Â âˆ ABCÂ  [Given, AB = AC]

2âˆ ABC = 1800Â – 480

2âˆ ABC = 1320

âˆ ABC = 660Â =Â âˆ ACB â€¦â€¦(i)

âˆ ACB = 660

âˆ ACD +Â âˆ DCB = 660

180Â +Â âˆ DCB = 660

âˆ DCB = 480Â â€¦â€¦â€¦(ii)

Now, InÂ âˆ†DCB,

âˆ DBC = 660Â  [From (i), SinceÂ âˆ ABC =Â âˆ DBC]

âˆ DCB = 480Â  [From (ii)]

âˆ BDC = 1800Â – 480Â – 660

âˆ BDC = 660

SinceÂ âˆ BDC =Â âˆ DBC

Therefore,Â BC = CD

Equal angles have equal sides opposite to them.

2. Calculate:

(ii)Â âˆ ABC

(iii)Â âˆ BAC

Solution:

Given: âˆ ACE = 1300; AD = BD = CD

Proof:

(i) âˆ ACD + âˆ ACE = 180o [DCE is a straight line]

âˆ ACD = 180o â€“ 130o

âˆ ACD = 50o

Now,

âˆ ACD = âˆ DAC = 50o â€¦ (i) [Since angles opposite to equal sides are equal]

âˆ ACD = âˆ DAC = 50o

âˆ ACD + âˆ DAC + âˆ ADC = 180o

50o + 50o + âˆ ADC = 180o

âˆ ADC = 180o â€“ 100o

(ii) âˆ ADC = âˆ ABD + âˆ DAB [Exterior angle is equal to sum of opposite interior angles]

âˆ´ âˆ DAB=âˆ ABD

80o = âˆ ABD + âˆ ABD

2âˆ BD = 80O

âˆ ABD = 40O = âˆ DAB â€¦ (ii)

(iii) We have,

âˆ BAC = âˆ DAB + âˆ DAC

Substituting the values from (i) and (ii),

âˆ BAC = 40O + 50O

Hence, âˆ BAC = 90O

3. In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate:

(i)Â âˆ CDE

(ii)Â âˆ DCE

Solution:

Given, âˆ FAB = 128O

âˆ BAC + âˆ FAB = 180O [As FAC is a straight line]

âˆ BAC = 180O â€“ 128O

âˆ BAC = 52O

In âˆ†ABC, we have

âˆ A = 52O

âˆ B = âˆ C [Given AB = AC and angles opposite to equal sides are equal]

Now, by angle sum property

âˆ A + âˆ B + âˆ C =180O

âˆ A + âˆ B + âˆ B = 180O

52O+ 2âˆ B = 180O

2âˆ B = 128O

âˆ B = 64O = âˆ Câ€¦ (i)

âˆ B = âˆ ADE [Given DE ll BC]

(i) Now, âˆ ADE + âˆ CDE + âˆ B = 180O [As ADB is a straight line]

64O + âˆ CDE + 64O= 180O

âˆ CDE = 180O+ 128O

âˆ CDE = 52O

(ii) Given DE ll BC and DC is the transversal

âˆ CDE = âˆ DCB = 52oâ€¦ (ii)

Also, âˆ ECB = 64oâ€¦ [From (i)]

But,

âˆ ECB = âˆ DCE +âˆ DCB

64o = âˆ DCE + 52o

âˆ DCE = 64o – 52o

âˆ DCE = 12o

4. Calculate x:

Solution:

(i) Let the triangle be ABC and the altitude be AD.

In âˆ†ABD, we have

âˆ DBA = âˆ DAB = 37o [Given BD = AD and angles opposite to equal sides are equal]

Now,

âˆ CDA = âˆ DBA + âˆ DAB [Exterior angle is equal to the sum of opposite interior angles]

âˆ CDA = 37o + 37o

âˆ´ âˆ CDA = 74o

âˆ CDA = âˆ CAD =Â 74o [Given CD = AC and angels opposite to equal sides are equal]

Now, by angle sum property

âˆ CAD + âˆ CDA + âˆ ACD = 180o

74o + 74o + x = 180o

x = 180o – 148o

x = 32o

(ii) Let triangle be ABC and altitude be AD.

In âˆ†ABD, we have

âˆ DBA = âˆ DAB = 50o [Given BD = AD and angles opposite to equal sides are equal]

Now,

âˆ CDA = âˆ DBA + âˆ DAB [Exterior angle is equal to the sum of opposite interior angles]

âˆ CDA = 50o + 50o

âˆ´ âˆ CDA = 100o

âˆ DAC = âˆ DCA =Â x [Given AD = DC and angels opposite to equal sides are equal]

So, by angle sum property

âˆ DAC + âˆ DCA + âˆ ADC = 180o

x + x + 100o = 180o

2x = 80o

x = 40o

5. In the figure, given below, AB = AC. Prove that:Â âˆ BOC =Â âˆ ACD.

Solution:

Letâ€™s assume âˆ ABO = âˆ OBC = x andÂ âˆ ACO =Â âˆ OCB = y

In ABC, we have

âˆ BAC = 180o – 2x – 2yâ€¦(i)

As, âˆ B = âˆ C [Since, AB = AC]

Â½ âˆ B = Â½ âˆ C

â‡’ x = y

Now,

âˆ ACD = 2x + âˆ BAC [Exterior angle is equal to sum of opposite interior angle]

= 2x + 180o – 2x – 2y [From (i)]

âˆ ACD = 180o – 2yâ€¦ (ii)

In âˆ†OBC, we have

âˆ BOC = 180o – x – y

âˆ BOC = 180o – y â€“ y [Since x = y]

âˆ BOC = 180o – 2yâ€¦ (iii)

Thus, from (ii) and (iii) we get

âˆ BOC = âˆ ACD

6. In the figure given below, LM = LN; âˆ PLN = 110o. Calculate:

(i)Â âˆ LMN

(ii)Â âˆ MLN

Solution:

Given, LM = LN and âˆ PLN = 110o

(i) We know that the sum of the measure of all the angles of a quadrilateral is 360o.

âˆ QPL + âˆ PLN +LNQ + âˆ NQP = 360o

90o + 110o + âˆ LNQ + 90o = 360o

âˆ LNQ = 360o â€“ 290o

âˆ LNQ = 70o

âˆ LNM = 70oâ€¦ (i)

In âˆ†LMN, we have

LM = LN [Given]

â‡’ âˆ LNM = âˆ LMN [ Angles opposite to equal sides are equal]

âˆ LMN = 70oâ€¦(ii) [From (i)]

(ii) In âˆ†LMN, we have

âˆ LMN + âˆ LNM + âˆ MLN = 180o

But, âˆ LNM = âˆ LMN = 70o [From (i) and (ii)]

â‡’ 70o + 70o + âˆ MLN = 180o

âˆ MLN = 180o – 140o

âˆ´ âˆ MLN = 40o

7. An isosceles triangle ABC has AC = BC. CD bisects AB at D andÂ âˆ CAB = 55o.

Find: (i)Â âˆ DCBÂ (ii)Â âˆ CBD.

Solution:

In âˆ†ABC, we have

AC = BC [Given]

So, âˆ CAB = âˆ CBD [Angles opposite to equal sides are equal]

â‡’ âˆ CBD = 55o

In âˆ†ABC, we have

âˆ CBA + âˆ CAB + âˆ ACB = 180o

But, âˆ CAB = âˆ CBA = 55o

55o + 55o + âˆ ACB = 180o

âˆ ACB = 180o – 110o

âˆ ACB = 70o

Now,

In âˆ†ACD and âˆ†BCD, we have

AC = BC [Given]

CD = CD [Common]

AD = BD [Given that CD bisects AB]

âˆ´ âˆ†ACD â‰… âˆ†BCD

So, By CPCT

âˆ DCA = âˆ DCB

âˆ DCB = âˆ ACB/2 = 70o/2

Thus, âˆ DCB = 35o

8. Find x:

Solution:

Letâ€™s put markings to the figure as following:

In âˆ†ABC, we have

âˆ´ âˆ ADC = âˆ ACD [Angles opposite to equal sides are equal]

Now,

âˆ ADC = âˆ DAB + DBA [Exterior angle is equal to the sum of opposite interior angles]

But,

âˆ DAB = âˆ DBA [Given: BD = DA]

âˆ´ âˆ ADC = 2âˆ DBA

2âˆ DBA = 42o

âˆ DBA = 21o

To find x:

x = âˆ CBA + âˆ BCA [Exterior angle is equal to the sum of opposite interior angles]

We know that,

âˆ CBA = 21o

âˆ BCA = 42o

â‡’ x = 21o + 42o

âˆ´ x = 63o

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 âˆ†ABC and âˆ†DBC, we have

BD = BD [Common]

âˆ BDA = âˆ BDC [Each equal to 90o]

âˆ ABD = âˆ DBC [BD bisects âˆ ABC]

âˆ´ âˆ†ABD â‰… âˆ†DBC [ASA criterion]

Therefore, by CPCT

x + 1 = y + 2

x = y + 1â€¦ (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 â€“ 2

3y + 4 = 5y â€“ 2

2y = 6

y = 3

Putting y = 3 in (i), we get

x = 3 + 1

âˆ´ x = 4

10. In the given figure; AE // BD, AC // ED and AB = AC. FindÂ âˆ a,Â âˆ b andÂ âˆ c.

Solution:

Letâ€™s assume points P and Q as shown below:

Given,Â âˆ PDQ = 58o

âˆ PDQ = âˆ EDC = 58o [Vertically opposite angles]

âˆ EDC = âˆ ACB = 58o [Corresponding angles âˆµ AC ll ED]

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ ACB = âˆ ABC = 58o [Angles opposite to equal sides are equal]

Now,

âˆ ACB + âˆ ABC + âˆ BAC = 180o

58o + 58o + a = 180o

âˆ a = 180o – 116o

âˆ a = 64o

Since, AE ll BD and AC is the transversal

âˆ ABC = âˆ b [Corrosponding angles]

âˆ´ âˆ b = 58o

Also, since AE ll BD and ED is the transversal

âˆ EDC = âˆ c [Corrosponding angles]

âˆ´ âˆ c = 58o

11. In the following figure; AC = CD, AD = BD andÂ âˆ C = 58o.

Find âˆ CAB.

Solution:

In âˆ†ACD, we have

AC = CD [Given]

âˆ´ âˆ CAD = âˆ CDA [Angles opposite to equal sides are equal]

And,

âˆ ACD = 58o [Given]

By angle sum property, we have

âˆ ACD + âˆ CDA + âˆ CAD = 180o

58o + 2âˆ CAD = 180o

âˆ CAD = âˆ CDA = 61oâ€¦ (i)

Now,

âˆ CDA = âˆ DAB + âˆ DBA [Exterior angles is equal to sum of opposite interior angles]

But,

âˆ DAB = âˆ DBA [Given, AD = DB]

So, âˆ DAB + âˆ DAB = âˆ CDA

2âˆ DAB = 61o

âˆ DAB = 30.5oâ€¦ (ii)

In âˆ†ABC, we have

âˆ CAB = âˆ CAD + âˆ DAB

âˆ CAB = 61o + 30.5o [From (i) and (ii)]

âˆ´ âˆ CAB = 91.5 o

12. In the figure of Q.11 is given above, if AC = AD = CD = BD; find angle ABC.

Solution:

In âˆ†ACD, we have

AC = AD = CD [Given]

Hence, ACD is an equilateral triangle

âˆ´ âˆ ACD = âˆ CDA = âˆ CAD = 60o

Now,

âˆ CDA = âˆ DAB + âˆ ABD [Exterior angle is equal to sum of opposite interior angles]

But,

âˆ DAB = âˆ ABD [Given, AD = DB]

So, âˆ ABD + âˆ ABD = âˆ CDA

2âˆ ABD = 60o

âˆ´ âˆ ABD = âˆ ABC = 30o

13. In âˆ†ABC; AB = AC andÂ âˆ A:Â âˆ B = 8: 5; find âˆ A.

Solution:

Let,Â âˆ A = 8x and âˆ B = 5x

Given, In âˆ†ABC

AB = AC

So, âˆ B = âˆ C = 5x [Angles opp. to equal sides are equal]

Now, by angle sum property

âˆ A + âˆ B +C = 180o

8x + 5x + 5x = 180o

18x = 180o

x = 10o

Thus, as âˆ A = 8x

âˆ A = 8 Ã— 10o

âˆ´ âˆ A = 80o

14. In triangle ABC;Â âˆ A = 60o,Â âˆ C = 40o, and bisector of angle ABC meets side AC at point P. Show that BP = CP.

Solution:

In âˆ†ABC, we have

âˆ A = 60o

âˆ C = 40o

âˆ´ âˆ B = 180o – 60o – 40o [By angle sum property]

âˆ B = 80o

Now, as BP is the bisector ofÂ âˆ ABC

âˆ´ âˆ PBC = âˆ ABC/2

âˆ PBC = 40o

In âˆ†PBC, we have

âˆ PBC = âˆ PCB = 40o

âˆ´ BP = CP [Sides opposite to equal angles are equal]

15. In triangle ABC; angle ABC = 90oÂ and P is a point on AC such thatÂ âˆ PBC =Â âˆ PCB. ShowÂ that:Â PA = PB.

Solution:

Letâ€™s assume âˆ PBC =Â âˆ PCB = x

In the right-angled triangle ABC,

âˆ ABC = 90o

âˆ ACB = x

âˆ BAC = 180o – (90o + x) [By angle sum property]

âˆ BAC = (90o – x) â€¦(i)

And

âˆ ABP = âˆ ABC – âˆ PBC

âˆ ABP = 90o â€“ x â€¦(ii)

Thus, in the âˆ†ABP from (i) and (ii), we have

âˆ BAP = âˆ ABP

Therefore, PA = PBÂ  [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, âˆ†ABC is an equilateral triangle

So, AB = BC = AC

âˆ ABC = âˆ CAB = âˆ ACB = 60o

Now, as sum of two non-adjacent interior angles of a triangle is equal to the exterior angle

âˆ CAB + âˆ CBA = âˆ ACE

60o + 60o = âˆ ACE

âˆ ACE = 120o

Now,

âˆ†ACE is an isosceles triangle with AC = CF

âˆ EAC = âˆ AEC

By angle sum property, we have

âˆ EAC + âˆ AEC + âˆ ACE = 180o

2âˆ AEC + 120o = 180o

2âˆ AEC = 180o – 120o

âˆ AEC = 30oÂ

Â

17. In triangle ABC, D is a point in AB such that AC = CD = DB. IfÂ âˆ B = 28Â°, find the angle ACD.

Solution:

From given, we get

âˆ†DBC is an isosceles triangle

â‡’ CD = DB

âˆ DBC = âˆ DCB [If two sides of a triangle are equal, then angles opposites to them are equal]

And, âˆ B = âˆ DBC = âˆ DCB = 28o

By angle sum property, we have

âˆ DCB + âˆ DBC + âˆ BCD = 180o

28o + 28o + âˆ BCD = 180o

âˆ BCD = 180o – 56o

âˆ BCD = 124o

As sum of two non-adjacent interior angles of a triangle is equal to the exterior angle, we have

âˆ DBC + âˆ DCB = âˆ DAC

28o + 28o = 56o

âˆ DAC = 56o

Now,

âˆ†ACD is an isosceles triangle with AC = DC

â‡’ âˆ ADC = âˆ DAC = 56o

âˆ ADC + âˆ DAC +âˆ DCA = 180o [By angle sum property]

56o + 56o + âˆ DCA = 180o

âˆ DCA = 180o – 112o

âˆ DCA = 64o

Thus, âˆ ACD = 64o

Â Â

18. In the given figure, AD = AB = AC, BD is parallel to CA and âˆ ACB = 65Â°. Find âˆ DAC.

Â

Solution:

From figure, itâ€™s seen that

âˆ†ABC is an isosceles triangle with AB = AC

â‡’ âˆ ACB = âˆ ABC

As âˆ ACB = 65o [Given]

âˆ´ âˆ ABC = 65o

By angle sum property, we have

âˆ ACB + âˆ CAB + âˆ ABC = 180o

65o + 65o + âˆ CAB = 180o

âˆ CAB = 180o – 130o

âˆ CAB = 50o

As BD is parallel to CA, we have

âˆ CAB = âˆ DBA as they are alternate angles

â‡’ âˆ CAB = âˆ DBA = 50o

Again, from figure, itâ€™s seen that

â‡’ âˆ ADB = âˆ DBA = 50o

By angle sum property, we have

âˆ ADB + âˆ DAB + âˆ DBA = 180o

50o + âˆ DAB + 50o = 180o

âˆ DAB = 180o – 100o

âˆ DAB = 80o

Now,

âˆ DAC = âˆ CAB + âˆ DAB

âˆ DAC = 50o – 80o

âˆ DAC = 130o

19. Prove that a triangle ABC is isosceles, if:

(i)Â Â altitude AD bisects angles BAC, or

(ii)Â Â bisector of angle BAC is perpendicular to base BC.

Solution:

(i)Â Â In
Î”ABC, if the altitude AD bisectÂ âˆ BAC.

Then, to prove: Î”ABC is isosceles.

In

So, by CPCT

AB = AC

Hence, Î”ABC is an isosceles.

(ii)Â Â In Î” ABC, the bisector ofÂ âˆ BAC is perpendicular to the base BC.

Then, to prove: Î”ABC is isosceles.

Â

Thus, by CPCT

AB = AC

Hence, Î”ABC is an isosceles.

20. In the given figure; AB = BC and AD = EC.

Prove that: BD = BE.

Â

Solution:

Â

In Î”ABC, we have

AB = BCÂ  (Given)

So, âˆ BCA =Â âˆ BACÂ  (Angles opposite to equal sides are equal)

â‡’Â âˆ BCD =Â âˆ BAEÂ â€¦.(i)

AD + DE = EC + DEÂ (Adding DE on both sides)

â‡’Â AE = CDÂ â€¦.(ii)

Now, in Î”ABE and Î”CBD, we have

AB = BCÂ  (Given)

âˆ BAE =Â âˆ BCDÂ  [From (i)]

AE = CDÂ  [From (ii)]

Therefore, Î”ABEÂ â‰…Â Î”CBD by SAS congruence criterion

So, by CPCT

BE = BD

Exercise 10(B)

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 âˆ DBCÂ and âˆ ECBÂ are formed.

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ C = âˆ B â€¦(i) [Angles opposite to equal sides are equal]

Since, âˆ B and âˆ C are acute they cannot be right angles or obtuse angles

Now,

âˆ ABC + âˆ DBC = 1800 [ABD is a straight line]

âˆ DBC = 1800 – âˆ ABC

âˆ DBC = 1800 – âˆ B â€¦(ii)

Similarly,

âˆ ACB + ECB = 1800 [ABD is a straight line]

âˆ ECB = 1800 – âˆ ACB

âˆ ECB = 1800 – âˆ C â€¦(iii)

âˆ ECB = 1800 – âˆ B â€¦(iv) [from (i) and (iii)]

âˆ DBC = âˆ ECB [from (ii) and (iv)]

Now,

âˆ DBC = 1800 – âˆ B

But, âˆ B is an acute angle

â‡’ âˆ DBC = 1800 â€“ (acute angle) = obtuse angle

Similarly,

âˆ ECB = 1800 – âˆ C

But, âˆ C is an acute angle

â‡’ âˆ ECB = 1800 â€“ (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

Solution:

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ C = âˆ B â€¦(i) [ Angles opposite to equal sides are equal]

(i) In âˆ†BPD and âˆ†CQD, we have

âˆ BPD = âˆ CQD [Each = 90o]

âˆ B = âˆ C [Proved]

BD = DC [Given]

Thus, âˆ†BPD â‰… âˆ†CQD by AAS congruence criterion

âˆ´ DP = DQ by CPCT

(ii) Since, âˆ†BPD â‰… âˆ†CQD

Therefore, BP = CQ [CPCT]

Now,

AB = AC [Given]

AB – BP = AC – CQ

AP = AQ

(iii) In âˆ†APD and âˆ†AQD, we have

DP = DQ [Proved]

AP = AQ [Proved]

Thus, âˆ†APD â‰… âˆ†AQD by SSS congruence criterion

3. In triangle ABC, AB = AC; BEÂ âŠ¥ AC and CFÂ âŠ¥ AB. Prove that:

(i) BE = CF

(ii) AF = AE

Solution:

(i) In âˆ†AEB and âˆ†AFC, we have

âˆ A = âˆ A [Common]

âˆ AEB = âˆ AFC = 90o [ Given : BE âŠ¥ AC and CE âŠ¥ AB]

AB = AC [Given]

Thus, âˆ†AEB â‰… âˆ†AFC by AAS congruence criterion

âˆ´ BE = CF by CPCT

(ii) Since, âˆ†AEB â‰… âˆ†AFC Â

âˆ ABE = âˆ AFC

âˆ´ AF= AE by CPCT

4. In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD. Prove that:Â âˆ BCD = 90o

Solution:

Construction: Join CD.

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ C = âˆ B â€¦ (i) [ Angles opposite to equal sides are equal]

In âˆ†ACD, we have

âˆ´ âˆ ADC = âˆ ACD â€¦ (ii)

Adding (i) and (ii), we get

âˆ B + âˆ ADC = âˆ C + âˆ ACD

âˆ B + âˆ ADC = âˆ BCD â€¦ (iii)

In âˆ†BCD, we have

âˆ B + âˆ ADC +âˆ BCD = 180o

âˆ BCD + âˆ BCD = 180o [From (iii)]

2âˆ BCD = 180o

âˆ´ âˆ BCD = 90o

5. (i) In âˆ†ABC, AB = AC andÂ âˆ A= 36Â°. If the internal bisector ofÂ âˆ 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 âˆ A = 36o

So, âˆ†ABC is an isosceles triangle.

âˆ B = âˆ C = (1800 â€“ 36o)/2 = 72o

âˆ ACD = âˆ BCD = 36o [âˆµ CD is the angle bisector of âˆ C]

Now, âˆ†ADC is an isosceles triangle as âˆ DAC = âˆ DCA = 36o

In âˆ†DCB, by angle sum property we have

âˆ CDB = 180o â€“ (âˆ DCB + âˆ DBC)

= 180o â€“ (36o + 72o)

= 180o – 108o

= 72o

Now, âˆ†DCB is an isosceles triangle as âˆ CDB = âˆ CBD = 72o

âˆ´ DC = BC â€¦(ii)

From (i) and (ii), we get

– Hence Proved.

6. Prove that the bisectors of the base angles of an isosceles triangle are equal.

Solution:

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ C = âˆ B â€¦(i) [Angles opposite to equal sides are equal]

Â½âˆ C = Â½âˆ B

â‡’ âˆ BCF = âˆ CBE â€¦(ii)

Now, in âˆ†BCE and âˆ†CBF, we have

âˆ C = âˆ B [From (i)]

âˆ BCF = âˆ CBE [From (ii)]

BC = BC [Common]

âˆ´ âˆ†BCE â‰… âˆ†CBF by AAS congruence criterion

Thus, BE = CF by CPCT

7. In the given figure, AB = AC andÂ âˆ DBC =Â âˆ ECB = 90o

Prove that:

(i) BD = CE

Solution:

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ ACB = âˆ ABC [Angles opposite to equal sides are equal]

â‡’ âˆ ABC = âˆ ACB â€¦ (i)

âˆ DBC =Â âˆ ECB = 90o [Given]

â‡’ âˆ DBC =Â âˆ ECB â€¦(ii)

Subtracting (i) from (ii), we get

âˆ DCB – âˆ ABC = âˆ ECB – âˆ ACB

âˆ DBA = âˆ ECA â€¦ (iii)

Now,

In Î”DBA and Î”ECA, we have

âˆ DBA = âˆ ECA [From (iii)]
âˆ DAB = âˆ EAC [Vertically opposite angles]
AB = AC [Given]
âˆ´ Î”DBA â‰… Î”ECA by ASA congruence criterion

Thus, by CPCT

BD = CE

And, also

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)Â âˆ BDA =Â âˆ CDA.

Solution:

DA is produced to meet BC in L

In âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ ACB = âˆ ABC â€¦ (i) [Angles opposite to equal sides are equal]

In âˆ†DBC, we have

DB = DC [Given]

âˆ´ âˆ DCB = âˆ DBC â€¦ (ii) [Angles opposite to equal sides are equal]

Subtracting (i) from (ii), we get

âˆ DCB – âˆ ACB = âˆ DBC – âˆ ABC

âˆ DCA = âˆ DBA â€¦(iii)

Now,

In âˆ†DBA and âˆ†DCA, we have

DB = DC [Given]

âˆ DBA = âˆ DCA [From (iii)]

AB = AC [Given]

âˆ´ âˆ†DBA â‰… âˆ†DCA by SAS congruence criterion

âˆ BDA = âˆ CDA â€¦(iv) [By CPCT]

In âˆ†DBA, we have

âˆ BAL = âˆ DBA + âˆ BDA â€¦(v) [Exterior angle = sum of opposite interior angles]

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

âˆ BAL = âˆ DCA + âˆ CDA â€¦(vi) [Exterior angle = sum of opposite interior angles]

In âˆ†DCA, we have

âˆ CAL = âˆ DCA + âˆ CDA â€¦(vi)

From (vi) and (vii)

âˆ BAL = âˆ CAL â€¦(viii)

In âˆ†BAL and âˆ†CAL,

âˆ BAL = âˆ CAL [From (viii)]

âˆ ABL = âˆ ACL [From (i)

AB = AC [Given]

âˆ´ âˆ†BAL â‰… âˆ†CAL by ASA congruence criterion

So, by CPCT

âˆ ALB = âˆ ALC

And, BL = LC â€¦(ix)

Now,

âˆ ALB + âˆ ALC = 180o

âˆ ALB + âˆ ALB = 180o [Using (ix)]

2âˆ ALB = 180o

âˆ ALB = 90o

âˆ´ AL âŠ¥ BC

Or DL âŠ¥ BC and BL âŠ¥ 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:

InÂ âˆ†ABC, we have AB = AC

âˆ B =Â âˆ C [Angles opposite to equal sides are equal]

Â½âˆ B = Â½âˆ C

âˆ OBC = âˆ OCB â€¦(i)

â‡’ OB = OC â€¦(ii) [Sides opposite to equal angles are equal]

Now,

InÂ âˆ†ABO andÂ âˆ†ACO, we have

AB = ACÂ  [Given]

âˆ OBC =Â âˆ OCBÂ  [From (i)]

OB = OCÂ  [From (ii)]

Thus, âˆ†ABO â‰… âˆ†ACO by SAS congruence criterion

So, by CPCT

âˆ BAO = âˆ CAO

Therefore, AO bisectsÂ âˆ BAC.

10. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.

Solution:

In âˆ†ABC, we have

AB = AC [Given]

âˆ C = âˆ B â€¦ (i) [Angles opposite to equal sides are equal]

Now,

Â½ AB = Â½ AC

BF = CE â€¦ (ii)

In âˆ†BCE and âˆ†CBF, we have

âˆ C = âˆ B [From (i)]

BF = CE [From (ii)]

BC = BC [Common]

âˆ´ âˆ†BCE â‰… âˆ†CBF by SAS congruence criterion

So, CPCT

BE = CF

11. Use the given figure to prove that, AB = AC.

Solution:

In âˆ†APQ, we have

AP = AQ [Given]

âˆ´ âˆ APQ = âˆ AQP â€¦(i) [Angles opposite to equal sides are equal]

In âˆ†ABP, we have

âˆ APQ = âˆ BAP + âˆ ABP â€¦(ii) [Exterior angle is equal to sum of opposite interior angles]

In âˆ†AQC, we have

âˆ AQP = âˆ CAQ + âˆ ACQ â€¦(iii) [Exterior angle is equal to sum of opposite interior angles]

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

âˆ BAP + âˆ ABP = âˆ CAQ + âˆ ACQ

But, âˆ BAP = âˆ CAQ [Given]

âˆ CAQ + ABP = âˆ CAQ + âˆ ACQ

âˆ ABP = âˆ CAQ + âˆ ACQ – âˆ CAQ

âˆ ABP = âˆ ACQ

âˆ B = âˆ C

So, in âˆ†ABC, we have

âˆ B = âˆ C

â‡’ 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

âˆ´ âˆ DAE = âˆ ABC = âˆ B [Corresponding angles]

Since, AE || BC and AC is the transversal

âˆ CAE = âˆ ACB = âˆ C [Alternate angles]

âˆ´ âˆ DAE = âˆ CAE

âˆ B = âˆ C

â‡’ 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) â€¦(i)Â

and AP = BQ = CR â€¦(ii)Â

Subtracting (ii) from (i), we get

AB – AP = BC – BQ = CA – CR

BP = CQ = AR â€¦(iii)

âˆ´ âˆ A = âˆ B = âˆ C â€¦(iv) [Angles opposite to equal sides are equal]

In âˆ†BPQ and âˆ†CQR, we have

BP = CQ [From (iii)]

âˆ B = âˆ C [From (iv)]

BQ = CR [Given]

âˆ´ âˆ†BPQ â‰… âˆ†CQR by SAS congruence criterion

So, PQ = QR [by CPCT] â€¦ (v)

In âˆ†CQR and âˆ†APR, we have

CQ = AR [From (iii)]

âˆ C = âˆ A [From (iv)]

CR = AP [Given]

âˆ´ âˆ†CQR â‰… âˆ†APR by SAS congruence criterion

So, QR = PR [By CPCT] â€¦ (vi)

From (v) and (vi), we get

PQ = QR = PR

Therefore, PQR is an equilateral triangle.

14. In triangle ABC, altitudes BE and CF are equal. Prove that the triangle is isosceles.

Solution:

InÂ âˆ†ABE andÂ âˆ†ACF, we have

âˆ A =Â âˆ A [Common]

âˆ AEB =Â âˆ AFC = 900 [Given: BEÂ âŠ¥ AC and CFÂ âŠ¥ AB]

BE = CF [Given]

âˆ´ âˆ†ABE â‰… âˆ†ACF by AAS congruence criterion

So, by CPCT

AB = AC

Therefore, ABC is an isosceles triangle.

15. Through any point in the bisector of angle, a straight line is drawn parallel to either arm of the angle. Prove that the triangle so formed is isosceles.

Solution:

Letâ€™s consider âˆ†ABC, AL is bisector of âˆ A.

Let D is any point on AL.

From D, a straight-line DE is drawn parallel to AC.

DE || AC Â [Given]

So, âˆ ADE =Â âˆ DAC …(i)Â  [Alternate angles]

âˆ DAC =Â âˆ DAE â€¦ (ii)Â  [AL is bisector ofÂ âˆ A]

From (i) and (ii), we get

âˆ´ AE = ED Â [Sides opposite to equal angles are equal]

Therefore, AED is an isosceles triangle.

16. In triangle ABC; AB = AC. P, Q and R are mid-points of sides AB, AC and BC respectively. Prove that:

(i) PR = QR (ii) BQ = CP

Solution:

(i)

InÂ âˆ†ABC, we have

AB = AC

Â½ AB = Â½ AC

AP = AQ … (i) [ Since P and Q are mid – points]

InÂ âˆ†BCA, we have

PR =Â Â½ AC [PR is line joining the mid – points of AB and BC]

PR = AQ â€¦ (ii)

InÂ âˆ†CAB, we have

QR =Â Â½ AB [QR is line joining the mid – points of AC and BC]

QR = AP â€¦(iii)

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

PR = QR

Â

(ii)

Given, AB = AC

â‡’ âˆ B =Â âˆ C

Also,

Â½ AB = Â½ AC

BP = CQ [P and Q are mid â€“ points of AB and AC]

Now, inÂ âˆ†BPC andÂ âˆ†CQB, we have

BP = CQ

âˆ B =Â âˆ C

BC = BC (Common)

Therefore,Â Î”BPC â‰… CQB by SAS congruence criterion

âˆ´ BP = CP by CPCT

17. From the following figure, prove that:

(i)Â âˆ ACD =Â âˆ CBE

Solution:

(i) InÂ âˆ†ACB, we have

AC = AC [Given]

âˆ´ âˆ ABC =Â âˆ ACB â€¦(i) [Angles opposite to equal sides are equal]

âˆ ACD +Â âˆ ACB = 1800 â€¦ (ii) [Since, DCB is a straight line]

âˆ ABC +Â âˆ CBE = 1800Â â€¦(iii) [Since, ABE is a straight line]

Equating (ii) and (iii), we get

âˆ ACD +Â âˆ ACB =Â âˆ ABC +Â âˆ CBE

âˆ ACD +Â âˆ ACB =Â âˆ ACB +Â âˆ CBE [From (i)]

â‡’ âˆ ACD =Â âˆ CBE

(ii) InÂ âˆ†ACD and âˆ†CBE, we have

DC = CB [Given]

AC = BE [Given]

âˆ ACD = âˆ CBE [Proved above]

âˆ´ âˆ†ACD â‰… âˆ†CBE by SAS congruence criterion

Hence, by CPCT

18. Equal sides AB and AC of an isosceles triangle ABC are produced. The bisectors of the exterior angle so formed meet at D. Prove that AD bisects angle A.

Solution:

AB is produced to E and AC is produced to F. BD is bisector of angle CBE and CD is bisector of angle BCF. BD and CD meet at D.

InÂ âˆ†ABC, we have

AB = AC [Given]

âˆ´ âˆ C =Â âˆ B [angles opposite to equal sides are equal]

âˆ CBE = 1800Â –Â âˆ B [ABE is a straight line]

âˆ CBD = (180o – âˆ B)/ 2 [BD is bisector of âˆ CBE]

âˆ CBD = 90o – âˆ B/ 2 â€¦(i)

Similarly,

âˆ BCF = 1800Â –Â âˆ C [ACF is a straight line]

âˆ BCD = (180o – âˆ C)/ 2 [CD is bisector of âˆ BCF]

âˆ BCD = 90o – âˆ C/2 â€¦(ii)

Now,

âˆ CBD = 90o – âˆ C/2 [âˆµâˆ B = âˆ C]

âˆ CBD = âˆ BCD

InÂ âˆ†BCD, we have

âˆ CBD = âˆ BCD

âˆ´ BD = CD

InÂ âˆ†ABD andÂ âˆ†ACD, we have

AB = AC [Given]

BD = CD [Proved]

âˆ´ âˆ†ABD â‰… âˆ†ACD by SSS congruence criterion

19. ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY. Prove thatÂ âˆ CAY =Â âˆ ABC.

Solution:

InÂ âˆ†ABC, we have

CX is the angle bisector ofÂ âˆ C

So, âˆ ACY =Â âˆ BCX â€¦(i)

InÂ âˆ AXY, we have

AX = AYÂ  [Given]

âˆ AXY =Â âˆ AYX …(ii)Â  [Angles opposite to equal sides are equal]

Now,Â âˆ XYC = âˆ AXB = 180Â°Â  [Straight line angle]

âˆ AYX +Â âˆ AYC =Â âˆ AXY +Â âˆ BXY

âˆ AYC =Â âˆ BXYâ€¦ (iii)Â  [From (ii)]

InÂ âˆ†AYC andÂ âˆ†BXC, we have

âˆ AYC +Â âˆ ACY +Â âˆ CAY =Â âˆ BXC +Â âˆ BCX +Â âˆ XBC = 180Â°

âˆ CAY =Â âˆ XBCÂ  [From (i) and (iii)]

Thus, âˆ CAY =Â âˆ ABC

20. In the following figure; IA and IB are bisectors of angles CAB and CBA respectively. CP is parallel to IA and CQ is parallel to IB.

Prove that:

PQ = The perimeter of theÂ âˆ†ABC.

Solution:

Since IA || CP and CA is a transversal

We have, âˆ CAI =Â âˆ PCA Â [Alternate angles]

Also, IA || CP and AP is a transversal

We have, âˆ IAB =Â âˆ APCÂ  [Corresponding angles]

ButÂ âˆ´Â âˆ CAI =Â âˆ IABÂ  [Given]

âˆ´Â âˆ PCA =Â âˆ APC

AC = AP

Similarly, BC = BQ

Now,

PQ = AP + AB + BQ

= AC + AB + BC

= Perimeter ofÂ âˆ†ABC

21. Sides AB and AC of a triangle ABC are equal. BC is produced through C upto a point D such that AC = CD. D and A are joined and produced upto point E. If angle BAE = 108o; find angle ADB.

Solution:

InÂ âˆ†ABD, we have

âˆ BAE =Â âˆ 3 +Â âˆ ADB

1080Â =Â âˆ 3 +Â âˆ ADB

But, AB = AC

âˆ 3 =Â âˆ 2

1080Â =Â âˆ 2 +Â âˆ ADB â€¦ (i)

Now,

InÂ âˆ†ACD, we have

âˆ 2 = âˆ 1 +Â âˆ ADB

But, AC = CD

Putting this value in (i), we get

22. The given figure shows an equilateral triangle ABC with each side 15 cm. Also, DE || BC, DF || AC and EG || AB. If DE + DF + EG = 20 cm, find FG.

Â Solution:

Given, ABC is an equilateral triangle.

AB = BC = AC = 15 cm

âˆ A = âˆ B = âˆ C = 60o

In âˆ†ADE, we have DE ll BC

âˆ AED = 60o [âˆµ âˆ ACB = 60o]

âˆ ADE = 60o [âˆµ âˆ ABC = 60o]

âˆ DAE = 180o â€“ (60o + 60o)

= 60o

Thus, âˆ†ADE is an equilateral triangle

Similarly, âˆ†BDF and âˆ†GEC are equilateral triangles

Now,

Let AD = x, AE = x and DE = x [âˆµ âˆ†ADE is an equilateral triangle]

Let BD = y, FD = y and FB = y [âˆµ âˆ†BDF is an equilateral triangle]

Let EC = z, GC = z and GE = z [âˆµ âˆ†GEC is an equilateral triangle]

Now, AD + DB = 15

x + y = 15 â€¦ (i)

AE + EC = 15

x + z = 15 â€¦ (ii)

Given, DE + DF + EG = 20

x + y + z = 20

15 + z = 20 [From (i)]

z = 5

From (ii), we get, x = 10

âˆ´ y = 5

Also, BC = 15

BF + FG + GC = 15

y + FG + z = 15

âˆ´ FG = 5

Â

23. If all the three altitudes of a triangle are equal, the triangle is equilateral. Prove it.

Solution:

In rightÂ âˆ†BEC andÂ âˆ†BFC, we have

BE = CF [Given]

BC = BC [Common]

âˆ BEC =Â âˆ BFC [Each = 900]

âˆ´ âˆ†BEC â‰… âˆ†BFC by RHS congruence criterion

By CPCT, we get

âˆ B = âˆ C

Similarly,

âˆ A = âˆ B

Hence, âˆ A = âˆ B = âˆ C

â‡’ AB = BC = AC

Therefore, ABC is an equilateral triangle.

24. In aÂ âˆ†ABC, the internal bisector of angle A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show thatÂ âˆ†ACE is isosceles.

Solution:

Given, DA || CE

âˆ 1 = âˆ 4 â€¦ (i) [Corresponding angles]

âˆ 2 = âˆ 3 â€¦ (ii) [Alternate angles]

ButÂ âˆ 1 = âˆ 2 â€¦(iii) [As AD is the bisector ofÂ âˆ A]

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

âˆ 3 = âˆ 4

â‡’AC = AE

Therefore, âˆ†ACE is an isosceles triangle.

25. In triangle ABC, bisector of angle BAC meets opposite side BC at point D. If BD = CD, prove thatÂ âˆ†ABC is isosceles.

Solution:

In âˆ†ABD and âˆ†EDC, we have

BD = CD [Given]

âˆ 1 = âˆ 2 [Vertically opposite angles]

âˆ´ âˆ†ABD â‰… âˆ†EDC by SAS congruence criterion

So, by CPCT,

AB = CE â€¦(i)

And, âˆ BAD = âˆ CED

âˆ´ âˆ CED = âˆ CAD

AC = CE â€¦(ii)

From (i) and (ii), we get

AB = AC

Hence, ABC is an isosceles triangle.

26. InÂ âˆ†ABC, D is point on BC such that AB = AD = BD = DC. Show that:

âˆ ADC:Â âˆ C = 4: 1.

Solution:

As, AB = AD = BD, we have

âˆ†ABD is an equilateral triangle.

Now,

= 180o – 60o

= 120o

âˆ´ âˆ 1 = âˆ 2

But,

âˆ 1 + âˆ 2 + âˆ ADC = 180o [By angle sum property]

2âˆ 1 + 120o = 180o

2âˆ 1 = 60o

âˆ 1 = 30o

âˆ C = 30o

â‡’âˆ ADC: âˆ C = 120o : 30o

Therefore, âˆ ADC: âˆ C = 4 : 1

27. Using the information given in each of the following figures, find the values ofÂ a andÂ b.Â Â [Given: CE = AC]

Solution:

(i) In âˆ†CAE, we have

âˆ CAE = âˆ AEC [âˆµ CE = AC]

= (180o â€“ 60o)/2

= 56o

In âˆ BEA, we have

a = 180o â€“ 56o = 124o

In âˆ†ABE, we have

âˆ ABE = 180o â€“ (124o + 14o)

= 180o – 138o

= 42oÂ

Â

(ii)

In âˆ†AEB and âˆ†CAD, we have

âˆ EAB = âˆ CAD [Given]

180o – âˆ ADE = 180o – âˆ AED

âˆ´ âˆ†AEB â‰… âˆ†CAD by ASA congruence criterion

Thus, AC = AB by CPCT

2a + 2 = 7b â€“ 1

2a â€“ 7b = – 3 â€¦ (i)

CD = EB

a = 3b â€¦ (ii)

Solving (i) and (ii) we get,

a = 9 and b = 3

Â

## Selina Solutions for Class 9 Maths Chapter 10-Isosceles Triangle

The Chapter 10, Isosceles Triangle, contains 2 exercises and the Selina Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

10.1 Introduction

10.2 Theorems:

1. If two sides of a triangle are equal, the angles opposite to them are also equal.
2. If two angles of a triangle are equal, the sides opposite to them are also equal.

## Selina Solutions for Class 9 Maths Chapter 10-Isosceles Triangle

An isosceles triangle is a triangle with (at least) two equal sides. This property is equal to two angles of the triangle being equal. Thus, we can say that an isosceles triangle has two equal sides as well as two equal angles. The Chapter 10 of class 9 gives the students an overview of the different properties and problems related to the Isosceles Triangle. Read and learn the Chapter 10 of Selina textbook to learn more about Isosceles Triangle along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.