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 = 48 ^{o}Â and**

**âˆ ACD = 18 ^{o}.**

**Show that BC = CD.**

**Solution:**

In âˆ†ABC, we have

**âˆ **BAC +Â **âˆ **ACB +Â **âˆ **ABC = 180^{0}

48^{0}Â +Â **âˆ **ACB +Â **âˆ **ABC = 180^{0}

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

2**âˆ **ABC = 180^{0}Â – 48^{0}

2**âˆ **ABC = 132^{0}

**âˆ **ABC = 66^{0}Â =Â **âˆ **ACB â€¦â€¦(i)

**âˆ **ACB = 66^{0}

**âˆ **ACD +Â **âˆ **DCB = 66^{0}

18^{0}Â +Â **âˆ **DCB = 66^{0}

**âˆ **DCB = 48^{0}Â â€¦â€¦â€¦(ii)

Now, InÂ âˆ†DCB,

**âˆ **DBC = 66^{0}Â [From (i), SinceÂ **âˆ **ABC =Â **âˆ **DBC]

**âˆ **DCB = 48^{0Â }[From (ii)]

**âˆ **BDC = 180^{0}Â – 48^{0}Â – 66^{0}

**âˆ **BDC = 66^{0}

SinceÂ **âˆ **BDC =Â **âˆ **DBC

Therefore,Â BC = CD

Equal angles have equal sides opposite to them.

**2. Calculate:**

**(i) âˆ ADC**

**(ii)Â âˆ ABC**

**(iii)Â âˆ BAC**

**Solution:**

Given: âˆ ACE = 130^{0}; AD = BD = CD

Proof:

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

âˆ ACD = 180^{o} â€“ 130^{o}

âˆ ACD = 50^{o}

Now,

CD = AD

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

In âˆ†ADC,

âˆ ACD = âˆ DAC = 50^{o}

âˆ ACD + âˆ DAC + âˆ ADC = 180^{o}

50^{o} + 50^{o} + âˆ ADC = 180^{o}

âˆ ADC = 180^{o} â€“ 100^{o}

âˆ ADC = 80^{o}

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

But, AD = BD

âˆ´ âˆ DAB=âˆ ABD

80^{o }= âˆ ABD + âˆ ABD

2âˆ BD = 80^{O}

âˆ ABD = 40^{O} = âˆ DAB â€¦ (ii)

(iii) We have,

âˆ BAC = âˆ DAB + âˆ DAC

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

âˆ BAC = 40^{O} + 50^{O}

Hence, âˆ BAC = 90^{O}

**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 = 128^{O}

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

âˆ BAC = 180^{O} â€“ 128^{O}

âˆ BAC = 52^{O}

In** **âˆ†ABC, we have

âˆ A = 52^{O}

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 =180^{O}

âˆ A + âˆ B + âˆ B = 180^{O}

52^{O}+ 2âˆ B = 180^{O}

2âˆ B = 128^{O}

âˆ B = 64^{O} = âˆ Câ€¦ (i)

âˆ B = âˆ ADE [Given DE ll BC]

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

64^{O }+ âˆ CDE + 64^{O}= 180^{O}

âˆ CDE = 180^{O}+ 128^{O}

âˆ CDE = 52^{O}

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

âˆ CDE = âˆ DCB = 52^{o}â€¦ (ii)

Also, âˆ ECB = 64^{o}â€¦ [From (i)]

But,

âˆ ECB = âˆ DCE +âˆ DCB

64^{o }= âˆ DCE + 52^{o}

âˆ DCE = 64^{o }– 52^{o }

âˆ DCE = 12^{o}

**4. Calculate x:**

** **

**Solution:**

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

In âˆ†ABD, we have

âˆ DBA = âˆ DAB = 37^{o} [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 = 37^{o} + 37^{o}

âˆ´ âˆ CDA = 74^{o}

Now, in âˆ†ADC, we have

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

Now, by angle sum property

âˆ CAD + âˆ CDA + âˆ 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 âˆ†ABD, we have

âˆ DBA = âˆ DAB = 50^{o} [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 = 50^{o} + 50^{o}

âˆ´ âˆ CDA = 100^{o}

In âˆ†ADC, we have

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 = 180^{o}

x + x + 100^{o} = 180^{o}

2x = 80^{o}

x = 40^{o}

**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 = 180^{o }– 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 + 180^{o }– 2x – 2y [From (i)]

âˆ ACD = 180^{o }– 2yâ€¦ (ii)

In âˆ†OBC, we have

âˆ BOC = 180^{o }– x – y

âˆ BOC = 180^{o }– y â€“ y [Since x = y]

âˆ BOC = 180^{o }– 2yâ€¦ (iii)

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

âˆ BOC = âˆ ACD

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

**(i)Â âˆ LMN**

**(ii)Â âˆ MLN**

**Solution:**

Given, LM = LN and âˆ 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,

âˆ QPL + âˆ PLN +LNQ + âˆ NQP = 360^{o}

90^{o} + 110^{o} + âˆ LNQ + 90^{o} = 360^{o}

âˆ LNQ = 360^{o} â€“ 290^{o}

âˆ LNQ = 70^{o}

âˆ LNM = 70^{o}â€¦ (i)

In âˆ†LMN, we have

LM = LN [Given]

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

âˆ LMN = 70^{o}â€¦(ii) [From (i)]

(ii) In âˆ†LMN, we have

âˆ LMN + âˆ LNM + âˆ MLN = 180^{o}

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

â‡’ 70^{o }+ 70^{o }+ âˆ MLN = 180^{o}

âˆ MLN = 180^{o} – 140^{o}

âˆ´ âˆ MLN = 40^{o}

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

**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 = 55^{o}

In âˆ†ABC, we have

âˆ CBA + âˆ CAB + âˆ ACB = 180^{o}

But, âˆ CAB = âˆ CBA = 55^{o}

55^{o} + 55^{o} + âˆ ACB = 180^{o}

âˆ ACB = 180^{o }– 110^{o}

âˆ ACB = 70^{o}

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 = 70^{o}/2

Thus, âˆ DCB = 35^{o}

**8. Find x:**

**Solution:**

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

In âˆ†ABC, we have

AD = AC [Given]

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

So, âˆ ADC = 42^{o}

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 = 42^{o}

âˆ DBA = 21^{o}

To find x:

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

We know that,

âˆ CBA = 21^{o}

âˆ BCA = 42^{o}

â‡’ x = 21^{o} + 42^{o}

âˆ´ 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 âˆ†ABC and âˆ†DBC, we have

BD = BD [Common]

âˆ BDA = âˆ BDC [Each equal to 90^{o}]

âˆ ABD = âˆ DBC [BD bisects âˆ ABC]

âˆ´ âˆ†ABD â‰… âˆ†DBC [ASA criterion]

Therefore, by CPCT

AD = DC

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 = 58^{o}

âˆ PDQ = âˆ EDC = 58^{o} [Vertically opposite angles]

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

In âˆ†ABC, we have

AB = AC [Given]

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

Now,

âˆ ACB + âˆ ABC + âˆ BAC = 180^{o}

58^{o} + 58^{o} + a = 180^{o}

âˆ a = 180^{o }– 116^{o}

âˆ a = 64^{o}

Since, AE ll BD and AC is the transversal

âˆ ABC = âˆ b [Corrosponding angles]

âˆ´ âˆ b = 58^{o}

Also, since AE ll BD and ED is the transversal

âˆ EDC = âˆ c [Corrosponding angles]

âˆ´ âˆ c = 58^{o}

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

**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 = 58^{o } [Given]

By angle sum property, we have

âˆ ACD + âˆ CDA + âˆ CAD = 180^{o}

58^{o} + 2âˆ CAD = 180^{o}

2âˆ CAD = 122^{o}

âˆ CAD = âˆ CDA = 61^{o}â€¦ (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 = 61^{o}

âˆ DAB = 30.5^{o}â€¦ (ii)

In âˆ†ABC, we have

âˆ CAB = âˆ CAD + âˆ DAB

âˆ CAB = 61^{o} + 30.5^{o} [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 = 60^{o}

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 = 60^{o}

âˆ´ âˆ ABD = âˆ ABC = 30^{o}

**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 = 180^{o}

8x + 5x + 5x = 180^{o}

18x = 180^{o}

x = 10^{o}

Thus, as âˆ A = 8x

âˆ A = 8 Ã— 10^{o}

âˆ´ âˆ A = 80^{o}

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

**Solution:**

In âˆ†ABC, we have

âˆ A = 60^{o}

âˆ C = 40^{o}

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

âˆ B = 80^{o}

Now, as BP is the bisector ofÂ âˆ ABC

âˆ´ âˆ PBC = âˆ ABC/2

âˆ PBC = 40^{o}

In âˆ†PBC, we have

âˆ PBC = âˆ PCB = 40^{o}

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

**15. In triangle ABC; angle ABC = 90 ^{o}Â 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 = 90^{o}

âˆ ACB = x

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

âˆ BAC = (90^{o} – x) â€¦(i)

And

âˆ ABP = âˆ ABC – âˆ PBC

âˆ ABP = 90^{o} 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 = 60^{o}

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

âˆ CAB + âˆ CBA = âˆ ACE

60^{o }+ 60^{o }= âˆ ACE

âˆ ACE = 120^{o}

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 = 180^{o}

2âˆ AEC + 120^{o} = 180^{o}

2âˆ AEC = 180^{o }– 120^{o}

âˆ AEC = 30^{o}**Â **

**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 = 28^{o}

By angle sum property, we have

âˆ DCB + âˆ DBC + âˆ BCD = 180^{o}

28^{o }+ 28^{o }+ âˆ BCD = 180^{o}

âˆ BCD = 180^{o} – 56^{o}

âˆ BCD = 124^{o}

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

28^{o }+ 28^{o }= 56^{o}

âˆ DAC^{ }= 56^{o}

Now,

âˆ†ACD is an isosceles triangle with AC = DC

â‡’ âˆ ADC = âˆ DAC = 56^{o}

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

56^{o} + 56^{o} + âˆ DCA = 180^{o }

âˆ DCA = 180^{o} – 112^{o}

âˆ DCA = 64^{o}

Thus, âˆ ACD = 64^{o}

**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 = 65^{o} [Given]

âˆ´ âˆ ABC = 65^{o}

By angle sum property, we have

âˆ ACB + âˆ CAB + âˆ ABC = 180^{o}

65^{o} + 65^{o} + âˆ CAB = 180^{o}

âˆ CAB = 180^{o} – 130^{o}

âˆ CAB = 50^{o}

As BD is parallel to CA, we have

âˆ CAB = âˆ DBA as they are alternate angles

â‡’ âˆ CAB = âˆ DBA = 50^{o}

Again, from figure, itâ€™s seen that

âˆ†ADB is an isosceles triangle with AD = AB.

â‡’ âˆ ADB = âˆ DBA = 50^{o}

By angle sum property, we have

âˆ ADB + âˆ DAB + âˆ DBA = 180^{o}

50^{o} + âˆ DAB + 50^{o} = 180^{o}

âˆ DAB = 180^{o }– 100^{o }

âˆ DAB = 80^{o}

Now,

âˆ DAC = âˆ CAB + âˆ DAB

âˆ DAC = 50^{o }– 80^{o}

âˆ DAC = 130^{o}

**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

Î”ADB and Î”ADC, we have

âˆ BAD =Â âˆ CADÂ (AD is bisector ofÂ âˆ BAC)

AD = ADÂ (Common)

âˆ ADB =Â âˆ ADCÂ (Each equal to 90Â°)

Therefore,Â Î”ADBÂ â‰…Â Î”ADCÂ by ASA congruence criterion

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 **

Â In Î”ADB and Î”ADC,

âˆ BAD =Â âˆ CADÂ (AD is bisector ofÂ âˆ BAC)

AD = ADÂ (Common)

âˆ ADB =Â âˆ ADCÂ (Each equal to 90Â°)

Therefore, Î”ADBÂ â‰…Â Î”ADCÂ by ASA congruence criterion

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)

Given, AD = EC

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 = 180^{0} [ABD is a straight line]

âˆ DBC = 180^{0} – âˆ ABC

âˆ DBC = 180^{0 }– âˆ B â€¦(ii)

Similarly,

âˆ ACB + ECB = 180^{0} [ABD is a straight line]

âˆ ECB = 180^{0} – âˆ ACB

âˆ ECB = 180^{0 }– âˆ C â€¦(iii)

âˆ ECB = 180^{0 }– âˆ B â€¦(iv) [from (i) and (iii)]

âˆ DBC = âˆ ECB [from (ii) and (iv)]

Now,

âˆ DBC = 180^{0 }– âˆ B

But, âˆ B is an acute angle

â‡’ âˆ DBC = 180^{0 }â€“ (acute angle) = obtuse angle

Similarly,

âˆ ECB = 180^{0 }– âˆ C

But, âˆ C is an acute angle

â‡’ âˆ ECB = 180^{0 }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**

**(iii) AD bisects âˆ A**

**Solution:**

Construction: Join AD.

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 = 90^{o}]

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]

AD = AD [Common]

AP = AQ [Proved]

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

âˆ PAD = âˆ QAD by CPCT

Hence, AD bisects angle A.

**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 = 90^{o } [ 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 = 90 ^{o}**

**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

AC = AD [Given]

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 = 180^{o}

âˆ BCD + âˆ BCD = 180^{o }[From (iii)]

2âˆ BCD = 180^{o}

âˆ´ âˆ BCD = 90^{o}

**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 = 36^{o}

So, âˆ†ABC is an isosceles triangle.

âˆ B = âˆ C = (180^{0} â€“ 36^{o})/2 = 72^{o}

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

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

âˆ´ AD = CD â€¦(i)

In âˆ†DCB, by angle sum property we have

âˆ CDB = 180^{o }â€“ (âˆ DCB + âˆ DBC)

= 180^{o} â€“ (36^{o }+ 72^{o})

= 180^{o }– 108^{o}

= 72^{o}

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

âˆ´ DC = BC â€¦(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 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 = 90 ^{o}**

**Prove that:**

**(i) BD = CE**

**(ii) AD = AE**

**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 = 90^{o }[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

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)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 = 180^{o}

âˆ ALB + âˆ ALB = 180^{o} [Using (ix)]

2âˆ ALB = 180^{o}

âˆ ALB = 90^{o}

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]

But, AE bisects âˆ CAD**Â **

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 = 90^{0 }[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 ADE =Â âˆ DAE

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**

**(ii) AD = CE**

**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 = 180^{0} â€¦ (ii) [Since, DCB is a straight line]

âˆ ABC +Â âˆ CBE = 180^{0}Â 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

AD = CE

**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 = 180^{0}Â –Â âˆ B [ABE is a straight line]

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

âˆ CBD = 90^{o} – âˆ B/ 2 â€¦(i)

Similarly,

âˆ BCF = 180^{0}Â –Â âˆ C [ACF is a straight line]

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

âˆ BCD = 90^{o} – âˆ C/2 â€¦(ii)

Now,

âˆ CBD = 90^{o} – 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]

AD = AD [Common]

BD = CD [Proved]

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

So, âˆ BAD = âˆ CAD [By CPCT]

Therefore, AD bisects âˆ A.

**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 = 108 ^{o}; find angle ADB.**

**Solution:**

InÂ âˆ†ABD, we have

âˆ BAE =Â âˆ 3 +Â âˆ ADB

108^{0}Â =Â âˆ 3 +Â âˆ ADB

But, AB = AC

âˆ 3 =Â âˆ 2

108^{0}Â =Â âˆ 2 +Â âˆ ADB â€¦ (i)

Now,

InÂ âˆ†ACD, we have

âˆ 2 = âˆ 1 +Â âˆ ADB

But, AC = CD

âˆ 1 =Â âˆ ADB

âˆ 2 =Â âˆ ADB +Â âˆ ADB

âˆ 2 = 2âˆ ADB

Putting this value in (i), we get

108^{0}Â = 2âˆ ADB +Â âˆ ADB

3âˆ ADB = 108^{0}

âˆ´ âˆ ADB = 36^{0}

**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 = 60^{o}

In âˆ†ADE, we have DE ll BC

âˆ AED = 60^{o } [âˆµ âˆ ACB = 60^{o}]

âˆ ADE = 60^{o } [âˆµ âˆ ABC = 60^{o}]

âˆ DAE = 180^{o }â€“ (60^{o }+ 60^{o})

= 60^{o}

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 = 90^{0}]

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:**

Letâ€™s produce AD up to E such that AD = DE.

In âˆ†ABD and âˆ†EDC, we have

AD = DE [By construction]

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

But, âˆ BAD = âˆ CAD [AD is bisector of âˆ BAC]

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.

âˆ´ âˆ ADB = 60^{o}

Now,

âˆ ADC = 180^{o }– âˆ ADB

= 180^{o }– 60^{o}

^{ }= 120^{o}

Again in âˆ†ADC, we have

AD = DC

âˆ´ âˆ 1 = âˆ 2

But,

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

2âˆ 1 + 120^{o} = 180^{o}

2âˆ 1 = 60^{o}

âˆ 1 = 30^{o}

âˆ C = 30^{o}

â‡’âˆ ADC: âˆ C = 120^{o} : 30^{o}

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]

= (180^{o} â€“ 60^{o})/2

= 56^{o }

In âˆ BEA, we have

a = 180^{o} â€“ 56^{o }= 124^{o}

In âˆ†ABE, we have

âˆ ABE = 180^{o} â€“ (124^{o} + 14^{o})

= 180^{o} – 138^{o}

= 42^{o}**Â **

**Â **

(ii)

In âˆ†AEB and âˆ†CAD, we have

âˆ EAB = âˆ CAD [Given]

âˆ ADC = âˆ AEB [âˆµ âˆ ADE = âˆ AED, since, AE = AD

180^{o }– âˆ ADE = 180^{o }– âˆ AED

âˆ ADC = âˆ AEB]

AE = AD [Given]

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:

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