Chapter 10 Isosceles Triangle

Exercise Ex. 10.1

1. In the figure alongside,

AB = AC

∠A = 48o and

∠ACD = 18o.

Show that BC = CD.

Solution

In ∆ ABC,

∠BAC + ∠ACB + ∠ABC = 180°

48° + ∠ACB + ∠ABC = 180°

But ∠ACB = ∠ABC [AB = AC]

2∠ABC = 180° - 48°

2∠ABC = 132°

∠ABC = 66° = ∠ACB ……(i)

∠ACB = 66°

∠ACD + ∠DCB = 66°

18° + ∠DCB = 66°

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∠DCB = 48°………(ii)

Now, In ∆ DCB,

∠DBC = 66° [From (i), Since ∠ABC = ∠DBC]

∠DCB = 48° [From (ii)]

∠BDC = 180° - 48° - 66°

∠BDC = 66°

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°; AD = BD = CD

Proof:

(i) ∠ACD +∠ACE =180° [ DCE is a st. Line ]

⇒ ∠ACD = 180° - 130°

⇒ ∠ACD = 50°

Now CD =AD

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⇒ ∠ACD = ∠DAC = 50° .... (i)

[Since angles opposite to equal sides are equal ]

In ∆ ADC,

∠ACD = ∠DAC = 50°

∠ACD + ∠DAC + ∠ADC = 180°

50° + 50° + ∠ADC = 180°

∠ADC = 180°- 100°

∠ADC = 80°

(ii) ∠ ADC = ∠ABD + ∠DAB [Exterior angle is equal to sum of opp.

Interior angle]

But AD = BD

∴ ∠DAB = ∠ABD

⇒ 80° = ∠ABD + ∠ABD

⇒ 2∠BD = 80°

⇒ ∠��� = 40° = ∠��� .... (ii)

(iii) ∠��� = ∠��� + ∠���

Substituting the value from (i) and (ii)

∠��� = 40° + 50°

⇒ ∠ ��� = 90°

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

BC. Calculate:

(i) ∠CDE

(ii) ∠DCE

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Solution

∠��� = 128° [Given]

∠��� + ∠��� = 180° [FAC IS st. Line]

⇒ ∠��� = 180° − 128°

⇒ ∠��� = 52°

In ∆���, ∠� = 52°

∠� = ∠� [Given AB = AC and angles opposite to equal sides are equal]

∠� + ∠� + ∠� = 180°

⇒ ∠� + ∠� + ∠� = 180° ⇒ 52° + 2∠� = 180°

⇒ ∠� = 64° = ∠� ...... (i)

∠� = ∠��� [Given DE ∥ BC]

(i) Now,

∠��� + ∠��� + ∠� = 180° [ADB is a st. Line]

⇒ 64° + ∠��� + 64° = 180°

⇒ ∠��� = 180° − 128° ⇒ ∠��� = 52°

(ii)

Given DE ∥ BC and DC is the transversal

⇒ ∠��� = ∠��� = 52° ..... (ii)

Also,∠��� = 64°.... [From (i)]

But,

∠��� = ∠��� + ∠���

⇒ 64° = ∠��� + 52°

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⇒ ∠��� = 64° − 52°

∠��� = 12°

4. Calculate x:

(i)

(ii)

Solution

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

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In ∆ ABD,

∠ ��� = ∠��� = 37° [Given BD =AD and angles opposite to equal

side are equal]

Now,

∠��� = ∠��� + ∠��� [Exterior angle is equal to the sum of opp.

Interior angles]

∴ ∠��� = 37° + 37°

⇒ ∠��� = 74°

Now in ∆ ADC,

∠��� = ∠��� = 74° [Given CD = AC and angles opposite to equal

sides are equal]

Now,

∠��� + ∠��� + ∠��� = 180°

⇒ 74° + 74° + = 180°

⇒ = 180° − 148° ⇒ = 32°

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

In ∆ ABD,

∠��� = ∠��� = 50° [Given BD = AD and angles opposite to equal

sides are equal]

Now,

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∠��� = ∠��� + ∠ ��� [Exterior angle is equal to the sum of opp.

Interior angles]

∴ ∠��� = 50° + 50°

⇒ ∠��� = 100°

In ∆ ADC,

∠ ��� = ∠��� = [Given AD =DC and angles opposite to equal sides

are equal]

∴ ∠ ��� + ∠��� + ∠��� = 180°

⇒ + + 100° = 180°

⇒ 2 = 80°

⇒ = 40°

5. in the figure, given below, AB = AC. Prove that: ∠BOC = ∠ACD.

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Solution

Let ∠ ��! = ∠!�� = "#$ ∠��! = ∠!�� = % In ∆ ABC,

∠��� = 180° − 2 − 2% ... (i)

Since ∠� = ∠� [AB =AC]

&' � = &

' �

⇒ = %

Now,

∠��� = 2 + ∠��� [Exterior angle is equal to sum of opp. Interior

angles]

= 2x +180° - 2x -2y [from (i)]

∠ ACD = 180° - 2Y......... (ii)

In ∆ OBC,

∠�!� = 180° − − %

⇒ ∠�!� = 180° − % − % [Already proved]

⇒ ∠�!� = 180° − 2y .....(iii)

From (i) and (ii)

∠�!� = ∠���

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

(i) ∠LMN

(ii) ∠MLN

Solution

Given: ∠PLN = 110 o

(i) We know that the sum of the measure of all the angles of a quadrilateral

is 360o.

In quad. PQNL,

∠ ()* + ∠)*+ + ∠*+( + ∠+() = 360°

⇒ 90° + 110° + ∠*+( + 90° = 360°

⇒ ∠*+( = 360° - 290°

⇒ ∠*+( = 70°

⇒ ∠*+, = 70° ........ (i)

In∠*,+ ,

LM = LN [Given]

∴ ∠*+, = ∠*,+ [angle opp. To equal sides are equal]

⇒ ∠*,+ = 70° ........ (ii) [From (i) ]

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-../ In ∆ LMN,

∠*,+ + ∠*+, + ∠,*+ = 180°

But, ∠ *+, = ∠*,+ = 70° [from (i) and (ii)]

∴ 70° + 70° + ∠,*+ = 180° ⇒ ∠,*+ = 180° − 140° ⇒ ∠,*+ = 40°

7. An isosceles triangle ABC has AC = BC. CD bisects AB at D and

∠CAB = 55o.

Find: (i) ∠DCB (ii) ∠CBD.

Solution

In ∆ ABC,

AC =BC [given]

∴ ∠��� = ∠��� [Angles opp. To equal sides are equal]

⇒ ∠��� = 55°

In ∆ ABC,

∠��� + ∠��� + ∠��� = 180°

But,

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∠��� = ∠��� = 55°

⇒ 55° + 55° + ∠��� = 180°

∠��� = 180° − 110° ⇒ ∠��� = 70°

Now,

In ∆ ACD and ∆ BCD,

AC= BC [given]

CD =CD [common]

AD =BD [Give: CD bisects AB]

∴ ∆ ACD ≅ ∆ BCD

⇒ ∠��� = ∠���

⇒ ∠��� = ∠123' = 45°

' ⇒ ∠��� = 35°

8. Find x:

Solution

Let us name the figure as following:

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In ∆ ABC,

AD = AC [Given]

∴ ∠ADC = ∠��� [angles opp. To equal sides are equal]

⇒ ∠ADC = 42°

Now,

∠ADC = ∠��� + ∠��� [Exterior angle is equal to the sum of opp.

Interior angles]

But,

∠��� = ∠��� [GIVEN: BD = DA]

∴ ∠ADC = 2∠���

⇒ 2∠��� = 42°

⇒ ∠��� = 21° For X:

X= ∠��� + ∠��� [Exterior angle is equal to the sum of opp. Interior

angles]

We know that,

∠��� = 21°

∠��� = 42°

∴ x = 21° + 42°

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⇒ x = 63°

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 ∆ABD and ∆ DBC,

BD = BD [common]

∠��� = ∠��� [Each equal to 90°]

∠��� = ∠��� [BD bisects ∠ABC]

∴ ∆ ABD ≅ ∆DBC [ASA criterion]

Therefore,

AD=DC

X +1 = y + 2

⇒ X = y +1..... (i)

And AB = BC

3x+1 = 5y-2

Substituting the value of x from (i)

3(y+1) +1 = 5y -2

⇒ 3y +3+1 = 5y -2

⇒ 3y + 4 = 5y -2

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⇒ 2y = 6

⇒ y =3

Putting y = 3 in (i)

x = 3 + 1

∴x = 4

10. In the given figure; AE // BD, AC // ED and AB = AC. Find ∠a, ∠b

and ∠c.

Solution

Let P and Q be the points as shown below:

Given: ∠ )�( = 58°

∠ PDQ = ∠ EDC = 58° [vertically opp. Angles]

∠ EDC =∠ACB =58° [Corresponding angles ∵ AC∥ED]

In ∆ ABC,

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AB = AC [Given]

∴ ∠ACB =∠ABC = 58° [angles opp. To equal sides are equal]

Now,

∠ ACB + ∠ABC +∠BAC = 180°

⇒ 58° + 58° + a = 180°

⇒ a =180° - 116°

⇒ a = 64°

Since AE ∥ �� and AC is the transversal

∠ABC = b [corresponding angles]

∴ b = 58°

Also since AE ∥ BD and ED is the transversal

∠ EDC = c [corresponding angles]

∴ c =58°

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

Find angle CAB.

Solution

In ∆ ACD,

AC = CD [given]

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∴ ∠CAD = ∠CDA

∠ ACD = 58o [given]

⟹ ∠58o + 2∠CAD = 180o

⟹ 2∠CAD = 122o

⟹ ∠CAD = ∠CDA = 61o.......... (i)

Now,

∠CDA = ∠DAB + ∠DBA [Ext. Angle is equal to sum of opp. Int.

Angles]

But,

∠DAB = ∠DBA [given: AD=DB]

∴ ∠DAB + ∠DAB = ∠CDA

⇒ 2∠DAB = 61o

⇒ ∠DAB = 30.5 o......... (ii)

In ∆ ABC,

∠CAB = ∠CAD +∠DAB

∴ ∠CAB = 61o + 30.5 o

⇒ ∠AB = 91.5 o

12. In the figure of q. no. 11 given above, if AC = AD = CD = BD; find

angle ABC.

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Solution

In ∆ACD,

AC =AD = CD [given]

Hence, ACD is an equilateral triangle

∴ ∠ACD =∠CDA =∠CAD = 60°

∠CDA = ∠DAB +∠ABD [Ext. Angle is equal to sum of opp. Int. Angles]

But,

∠DAB =∠ABD [given: AD=DB]

∴ ∠ABD +∠ABD =∠CDA

⇒ 2∠ABD = 60°

⇒ ∠ABD = ∠ABC = 30°

13. In triangle ABC; AB = AC and ∠A: ∠B = 8: 5; find angle A.

Solution

Let ∠A = 8x and ∠B = 5x

Given: AB = AC

⇒ ∠B = ∠C = 5x [Angles opp. to equal sides are equal]

Now,

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∠A + ∠B + ∠C = 180°

⇒ 8x + 5x + 5x = 180°

⇒ 18x = 180°

⇒ x = 10° Given that:

∠A = 8x

⇒ ∠A = 8 × 10°

⇒ ∠A = 80°

14. In triangle ABC; ∠A = 60o, ∠C = 40o, and bisector of angle ABC

meets side AC at point P. Show that BP = CP.

Solution

In ∆ABC,

∠A= 60°

∠C = 40°

∴ ∠B = 180° - 60° - 40°

⇒ ∠B = 80°

Now,

BP is the bisector of ∠ABC

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∴ ∠PBC = ∠132

'

⇒ ∠PBC = 40°

In ∆PBC

∠PBC = ∠PCB = 40°

∴ BP =CP [sides opp. To equal angles are equal]

15. In triangle ABC; angle ABC = 90o and P is a point on AC such

that ∠PBC = ∠PCB. Show that: PA = PB.

Solution

Let ∠PBC = ∠PCB = x

In the right angled triangle ABC,

∠ABC = 90°

∠ABC = x

⇒ ∠BAC = 180° - (90° + x)

⇒ ∠BAC = (90° - x)......... (i)

And

∠ABP = ∠ABC - ∠PBC

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⇒ ∠ABP = 90°- x......... (ii)

Therefore in the triangle ABP;

∠BAP = ∠ABP

Hence,

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

∆ ABC is an equilateral triangle

⇒ Side AB = side AC

⇒ ∠ABC = ∠ACB 8 if two sides of a triangle are equal , then angles opposite to them are equalK

Similarly, side AC = side BC

⇒ ∠CAB =∠ABC 8 if two sides of a triangle are equal , then angles opposite to them are equalK

Hence, ∠ABC = ∠CAB = ∠ABC = Y(say)

As the sum of all the angles of the triangle is 180°

∠ABC + ∠CAB + ∠ACB = 180°

⇒ 3Y = 180°

⇒ Y = 60°

∠ABC = ∠CAB =∠ACB = 60°

Sum of two non – adjacent interior angles of a triangle is equal to the

exterior angle

⇒ ∠CAB + ∠CBA = ∠ACE

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⇒ 60° + 60° = ∠ACE

⇒ ∠ACE = 120°

Now ∆ACE is an isosceles triangle with AC = CF

⇒ ∠EAC = ∠AEC

Sum of all the angles of a triangle is 180°

∠EAC+ ∠AEC + ∠ACE = 180°

⇒ 2∠AEC = 120° = 180°

⇒ ∠AEC = 30°

17. In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B

= 28°, find the angle ACD.

Solution

∆ DBC is an isosceles triangle

As, side CD = side DB

⇒ ∠DBC = ∠DCB 8 if two sides of a triangle are equal , then angles opposite to them are equalK

And ∠B = ∠DBC =∠DCB = 28°

As the sum of all the angles of the triangle is 180°

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∠DCB + ∠DBC + ∠BCD = 180°

⇒ 28° + 28° + ∠BCD = 180°

⇒ ∠BCD = 180° - 56°

⇒ ∠BCD = 124°

Sum of two non – adjacent interior angles of a triangle is equal to the

exterior angle.

⇒ ∠DBC + ∠DCB = ∠DAC

⇒ 28° + 28° = 56°

⇒ ∠DAC = 56°

Now ∆ACD is an isosceles triangle with AC = DC

⇒ ∠ADC = ∠DAC = 56°

Sum of all the angles of a triangle is 180°

⇒ 56°+ 56° + ∠DAC = 180°

⇒ ∠DAC = 180° - 112°

⇒ ∠DAC = 64° = ∠ADC

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18. In the given figure, AD = AB = AC, BD is parallel to CA and angle

ACB = 65°. Find angle DAC.

Solution

We can see that the ∆ABC is an isosoeles triangle with side AB = side AC

⇒ ∠ACB = ∠ABC

As ∠ACB = 65°

Hence ∠ABC = 65°

Sum of all the angles of a triangle is 180°

∠ACB+ ∠CAB + ∠ABC = 180°

65° + 65° + ∠CAB = 180°

∠CAB = 180°- 130°

∠CAB = 50°

As BD is parallel to CA

Therefore, ∠CAB =∠DBA since they are alternate angles.

∠CAB=∠DBA = 50°

We see that ∆ADB is an isosceles triangle with side AD = side AB.

⇒ ∠ADB = ∠DBA = 50°

Sum of all the angles of a triangle is 180°

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∠ADB + ∠DAB + ∠DBA = 180°

50° + ∠DBA + 50° = 180°

∠DBA = 180° - 100° = 80°

∠DBA= 80°

The angle DAC is sum of angle DAB and CAB.

∠DAC = ∠CAB + ∠DAB

∠DAC = 50° + 80°

∠DAC = 130°

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, let the altitude AD bisects ∠BAC.

Then we have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

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∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ AB = AC (cpct)

Hence, ΔABC is an isosceles.

(ii) In Δ ABC, the bisector of ∠ BAC is perpendicular to the base BC. We

have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ AB = AC (cpct)

Hence, ΔABC is an isosceles.

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

Prove that: BD = BE.

Solution

In ΔABC,

AB = BC (given)

⇒ ∠BCA = ∠BAC (Angles opposite to equal sides are equal)

⇒ ∠BCD = ∠BAE ….(i)

Given, AD = EC

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

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⇒ AE = CD ….(ii)

Now, in triangles ABE and CBD,

AB = BC (given)

∠BAE = ∠BCD [From (i)]

AE = CD [From (ii)]

⇒ ΔABE ≅ ΔCBD

⇒ BE = BD (cpct)

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Exercise 10.2

1. If the equal sides of an isosceles triangle are produced, prove that the

exterior angles so formed are obtuse and equal.

Solution

Const: AB is produced to D and AC is produced to E so that exterior angles

∠DBC and ∠ECB is formed.

In ∆ABC,

AB = AC [given]

∴ ∠C = ∠B...... (i) [Angles opp. To equal sides are equal]

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

angle.

∠ ABC + ∠ DBC = 180° [ABD is a st. Line]

⇒ ∠ DBC = 180°- ∠ ABC

⇒ ∠ DBC = 180° - ∠ B..... (ii)

Similarly,

∠ ACB + ∠ ECB = 180° [ABD is a st. Line]

⇒ ∠ ECB = 180° - ∠ ACB

⇒ ∠ ECB = 180° - ∠ C...... (iii)

⇒ ∠ ECB = 180° - ∠ B...... (iv) [from (i) and (ii) ]

⇒ ∠ DBC = ⇒ ∠ ECB [from (ii) and (iv)]

Now,

⇒ ∠ DBC = 180° - ∠ B

But ∠ B = Acute angle

∴ ∠ DBC = 180° - Acute angle = obtuse angle

Similarly,

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∠ ECB = 180° - ∠C

But ∠C = Acute angle

∴ ∠ECB = 180° - 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 angle A

Solution

Const: Join AD.

In ∆ABC,

AB = AC [given]

∴ ∠C = ∠B..... (i) [Angles opp. To equal sides are equal]

(i) in ∆BPD and ∆CQD,

∠ BPD = ∠CQD [earth = 90° ]

∠B = ∠C [proved]

BD = DC [given]

∴ ∆BPD ≅ ∆CQD [AAS criterion]

∴ DP = DQ [cpct]

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(ii) We have already proved that ∆BPD ≅ ∆CQD

Therefore, BP = CQ [cpct]

Now,

AB = AC [given]

⇒ AB – BP =AC - CQ

⇒ AP =AQ

(iii) In ∆APD and ∆AQD,

DP = DQ [proved]

AD = AD [common]

AP = AQ [Proved]

∴ ∆ APD ≅ ∆AQD [SSS]

⇒ ∠PAD = ∠QAD [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,

∠ A = ∠ A [common]

∠AEB = ∠AFC = 90° [given: BE⊥ AC]

[Given: CF ⊥ AB]

AB = AC [Given]

⇒ ∆AEB ≅ ∆AFC [AAS]

∴ BE = CF [cpct]

(ii)Since ∆AEB ≅ ∆AFC

∠ ABE = ∠AFC

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∴ AF = AE [congruent angles of congruent triangles]

4. In isosceles triangle ABC, AB = AC. The side BA is produced to D

such that BA = AD. Prove that: ∠BCD = 90o

Solution

Const: Join CD.

In ∆ ABC,

AB = AC [given]

∴ ∠C = ∠B..... (i) [Angles opp. to equal sides are equal]

In ∆ ACD,

AC = AD [Given]

∴ ∠ADC = ∠ACD..... (ii)

Adding (i) and (ii)

∠ B + ∠ADC = ∠C +∠ACD

∠ B + ∠ADC = ∠BCD..... (iii)

In ∆ BCD,

∠ B + ∠ADC + ∠BCD = 180°

∠BCD + ∠BCD = 180° [from (iii)]

2∠BCD = 180°

∠BCD = 90°

5. (i) In triangle ABC, AB = AC and ∠A = 36°. If the internal bisector

of ∠M 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.

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Solution

AC = AC

∆ABC is an isosceles triangle

∠A = 36°

∠B = ∠C = &N5°OPQ°

' = 72°

∠ACD = ∠BCD = 36° [∵CD is the angle bisector of ∠C]

∆ABC is an isosceles triangle since ∠DAC = ∠DCA = 36°

∴ AD = CD..... (i)

In ∆DCB,

∠CDB = 180° - (∠DCB + ∠DBC)

= 180° - (36° + 72°)

= 180° - 108°

= 72°

∆ DCB is an isosceles triangle since ∠CDB = ∠CBD = 72°

∴ 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

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In ∆ ABC,

AB =AC [given]

∴ ∠C = ∠B..... (i) [Angles opp. to equal sides are equal]

⇒ &' ∠C =

&' ∠B

⇒ ∠BCF = ∠CBE.... (ii)

In ∆BCE and ∆CBF,

∠C = ∠B [from (i)]

∠BCF = ∠CBF [from (ii)]

BC = BC [Common]

∴ ∆ BCE ≅ ∆CBF [AAS]

⇒ BE = CF [cpct]

7. In the given figure, AB = AC and ∠DBC = ∠ECB = 90o

Prove that:

(i) BD = CE

(ii) AD = AE

Solution

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In ∆ABC,

AB = AC [given]

∴ ∠ACB = ∠ABC [angles opp. to equal sides are equal]

⇒ ∠ABC = ∠ACB..... (i)

∠DBC = ∠ECB = 90o [Given]

⟹ ∠DBC = ∠ECB …….(ii)

Subtracting (i) from (ii)

∠DCB - ∠ABC = ∠ECB - ∠ACB

⇒ ∠DBA = ∠ECA.... (iii)

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,

AB =AC [given]

∴ ∠ACB = ∠ABC..... (i) [Angles opposite to equal sides are equal]

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In ∆DBC,

DB = DC [given]

∴ ∠ DCB = ∠ DBC..... (ii) [Angles opposite to equal sides are equal]

Subtracting (i) from (ii)

∠ DCB - ∠ACB = ∠DBC - ∠ABC

⇒ ∠ DBA = ∠DCA [ from (iii)]

AB = AC [Given]

∴ ∆ DBA ≅ ∆ DCA [SAS]

⇒ ∠BDA = ∠CDA........ (iv) [cpct]

In ∆DBA,

∠BAL = ∠DBA + ∠BDA....... (v)

[Ext. Angle = sum of opp. int. Angles]

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

∠BAL = ∠DCA + ∠CDA....... (vi)

In ∆DCA,

∠CAL = ∠DCA + ∠CDA...... (vii)

[Ext. Angle = sum of opp. int. Angles]

From (vi) and (viii)

∠BAL = ∠CAL...(viii)

In ∆ BAL and ∆CAL, [From (viii)]

∠ ABL = ∠ACL [from (i)]

AB = AC [given]

∴ ∆ BAL ≅ ∆CAL [ASA]

⇒ ∠ALB = ∠ALC [cpct]

And BL = LC....... (ix) [Cpct]

Now,

∠ALB + ∠ALC = 180°

⇒ ∠ALB + ∠ALB = 180°

⇒ 2∠ALB = 180°

⇒ ∠ALB = 90°

∴ AL ⊥ BC

Or DL ⊥ BC and BL = LC

∴ DA produced bisects BC at right angle.

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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]

⇒ &' ∠B =

&' ∠C

⇒ ∠OBC = ∠OBC........ (i)

⇒ OB = OC......... (ii)

[Angles opposite to equal sides are equal]

Now,

In ∆ABO and ∆ACO,

AB = AC [Given]

∠OBC = ∠OCB [From (i)]

OB = OC [From (ii)]

∆ ABO ≅ ∆ ACO [SAS criterion]

⇒ ∠BAO = ∠CAO [cpct]

Therefore, AO bisects ∠BAC.

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

triangle are equal.

Solution

In ∆ ABC,

AB = AC [given]

∴ ∠C = ∠B.... (i) [Angles opp. to equal sides is equal]

⇒ &'AB =

&'AC

⇒ BF = CE...... (ii)

⇒ &'AB =

&'AC

⇒ BF = CE....... (iii)

⇒ In ∆ BCE and ∆CBF,

∠C = ∠B [from (i) ]

BF = CE [ from (ii) ]

BC = BC [Common]

∴ ∆ BCE ≅ ∆CBF [SAS]

⇒ BE = CF [cpct]

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

Solution

In ∆APQ

AP = AQ [given]

∴ ∠APQ = ∠APQ........ (i)

[Angles opposite to equal sides are equal]

In ∆ABP,

∠ APQ = ∠CAQ + ∠ACQ......... (iii)

[Ext angle is equal to sum of opp. int. Angles]

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

∠BAP + ∠ABP = ∠CAQ +∠ACQ0

But, ∠BAP = ∠CAQ [given]

⇒ ∠CAQ + ∠ABP = ∠CAQ + ∠ACQ

⇒ ∠ABP = ∠CAQ + ∠ACQ - ∠CAQ

⇒ ∠ABP = ∠ACQ

⇒ ∠B = ∠C........ (iv)

In ∆ABC,

∠B = ∠C

⇒ AB = AC [sides opposite to equal angles are equal]

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12. In the given figure; AE bisects exterior angle CAD and AE is

parallel to BC.

Prove that: AB = AC.

Solution 12

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

⇒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

AB = BC = CA…….(i) [Given]

AP = BQ = CR…….(ii) [Given]

Subtracting (ii) from (i)

AB - AP = BC - BQ = CA - CR

BP = CQ = AR ………… (iii)

∴ ∠� = ∠� = ∠�…….. (iv) [Angles opp. to equal sides is equal]

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In ∆BPQ and ∆CQR,

BP = CQ [from (iii)]

∠B = ∠C [from (iv)]

BQ = CR [given]

∴ ∆BPQ ≅ ∆CQR [SAS criterion]

⇒ PQ = QR..... (v)

In ∆CQR and ∆APR,

CQ = AR [from (iii)]

∠C = ∠A [from (iv)]

CR = AP [given]

∴ ∆CQR ≅ ∆APR [SAS criterion]

⇒ QR = PR..... (vi)

From (v) and (vi)

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

In ∆ABE and ∆ACF,

∠A = ∠A [Common]

∠AEB = ∠AFC = 900[Given: BE ⊥AC; CF ⊥ AB]

BE = CF [Given]

∴ ∆ABE ≅ ∆ACF [AAS criterion]

⇒ AB = AC

Therefore, ABC is an isosceles triangle.

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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

AL is bisector of angle A. Let D is any point on AL. From D, a straight line

DE is drawn parallel to AC.

DE || AC [Given]

∴ ∠ADE = ∠DAC….(i) [Alternate angles]

∠DAC = ∠DAE…….(ii) [AL is bisector of A]

From (i) and (ii)

∠ADE = ∠DAE

∴AE = ED [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)

In ∆ABC,

AB = AC

⇒ &'AB =

&'AC

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

In ∆BCA,

PR = &'AC [PR is line joining the mid - points of AB and BC]

⇒PR = AQ…….. (ii)

In ∆CAB,

QR = &'AB [QR is line joining the mid - points of AC and BC]

⇒QR = AP…… (iii)

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

PR = QR

(ii)

AB = AC

⇒ ∠B = ∠C

Also, &' AB =

&' AC

⇒ BP = CQ [P and Q are mid – points of AB and AC]

In ∆BPC and ∆CQB,

BP = CQ

∠B = ∠C

BC = BC

Therefore, ΔBPC≅ΔCQB [SAS]

BP = CP

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17 .From the following figure, prove that:

(i) ∠ACD = ∠CBE

(ii) AD = CE

Solution

(i) In ∆ACB,

AC = AC [Given]

∴ ∠ABC = ∠ACB ……. (i)[Angles opposite to equal sides are equal]

∠ACD + ∠ACB = 1800 ……. (ii)[DCB is a straight line]

∠ABC + ∠CBE = 1800 …….. (iii)[ABE is a straight line]

Equating (ii) and (iii)

∠ACD + ∠ACB = ∠ABC + ∠CBE

⇒ ∠ACD + ∠ACB = ∠ACB + ∠CBE [From (i)]

⇒ ∠ACD = ∠CBE

(ii)

In ∆ ACD and ∆ CBE,

DC = CB [given]

AC =BE [given]

∠ACD = ∠CBE [proved earlier]

∴ ∆ ACD ≅ ∆CBE [SAS criterion]

⇒ AD = CE [cpct]

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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.

In ∆ABC,

AB = AC [Given]

∴ ∠C = ∠B[angles opposite to equal sides are equal]

∠CBE = 1800 - ∠B [ABE is a straight line]

⇒ ∠��� = &N5°O∠3' [BD is bisector of ∠CBE]

⇒ ∠��� = 90° − ∠3' ............ (i)

Similarly,

∠BCF = 1800 - ∠C [ACF is a straight line]

⇒ ∠��� = &N5°O∠2' [CD is bisector of ∠BCF]

⇒ ∠��� = 90° − ∠2' ............ (ii)

Now,

⇒ ∠CBD = 90° − ∠2' [∵ ∠B = ∠C]

In ∆BCD,

∠CBD = ∠BCD

∴BD = CD

In ∆ABD and ∆ACD,

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AB = AC [Given]

AD = AD [Common]

BD = CD [Proved]

∴ ∆ ABD ≅ ∆ ACD [SSS criterion]

Therefore, AD bisects ∠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 that ∠CAY = ∠ABC.

Solution

In ∆ABC,

CX is the angle bisector of ∠C

⇒ ∠ACY = ∠BCX....... (i)

In ∆AXY,

AX = AY [Given]

∠AXY = ∠AYX ……. (ii) [Angles opposite to equal sides are equal]

Now ∠XYC = ∠AXB = 180° [straight line]

∠AYX + ∠AYC = ∠AXY + ∠BXY

⇒ ∠AYC = ∠BXY........ (iii) [From (ii)]

In ∆AYC and ∆BXC

∠AYC + ∠ACY + ∠CAY = ∠BXC + ∠BCX + ∠XBC = 180°

⇒ ∠CAY = ∠XBC [From (i) and (iii)]

⇒ ∠CAY = ∠ABC

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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 the ∆ABC.

Solution Since IA || CP and CA is a transversal

∴ ∠CAI = ∠PCA [Alternate angles]

Also, IA || CP and AP is a transversal

∴ ∠IAB = ∠APC [Corresponding angles]

But ∠CAI = ∠IAB [Given]

∴ ∠PCA = ∠APC

⇒AC = AP

Similarly,

BC = BQ

Now,

PQ = AP + AB + BQ

= AC + AB + BC

= Perimeter of ∆ABC

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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

In ∆ABD,

∠BAE = ∠3 + ∠ADB

⇒1080 = ∠3 + ∠ADB

But AB = AC

⇒ ∠3 = ∠2

⇒1080 = ∠2 + ∠ADB …… (i)

Now,

In ∆ACD,

∠2=∠1+ ∠ADB

But AC = CD

⇒ ∠1 = ∠ADB

⇒ ∠2 = ∠ADB + ∠ADB

⇒ ∠2 = 2∠ADB

Putting this value in (i)

⇒1080 = 2∠ADB + ∠ADB

⇒3∠ADB = 1080

⇒ ∠ADB = 360

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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.

Solution ABC is an equilateral triangle.

Therefore, AB = BC = AC = 15 CM

∠A = ∠B = ∠C = 60°

In ∆ ADE, DE ∥ BC [given]

∠ AED = 60° [ ∵ ∠ACB = 60°]

∠ AED = 60° [ ∵ ∠ABC = 60°]

∠DAE = 180°- (60° + 60°) = 60°

Similarly, ∆ BDF and ∆GEC are equilateral triangles.

= 60°[∵ ∠C = 60°]

Let AD = x, AE = x, DE = x [∴ ∆ADE is an equilateral triangle]

Let BD = y, FD = y, FB = y [∵ ∆BDF is an equilateral triangle]

Let EC = z, GC = z, GE = z [∵ ∆GEC is an equilateral triangle]

Now, AD + DB =15 ⇒ x + y = 15...... (i)

AE + EC = 15 ⇒ x + y = 15....... (ii)

Given, DE + DF + EG = 20

⇒ x + y + z =20

⇒ 15 + z = 20 [from (i)]

⇒ z = 5

From (ii) we get x = 10

∴ y = 5

Also, BC = 15

BF + FG + GC = 15

⇒ y + FG + Z = 15

⇒ 5 + FC + 5 = 15

⇒ FC = 5.

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

equilateral. Prove it.

Solution

In right ∆BEC and ∆BFC,

BE = CF [Given]

BC = BC [Common]

∠BEC = ∠BFC [each = 900]

∴ ∆ BEC ≅ ∆ BFC [RHS]

⇒ ∠B = ∠C

Similarly,

∠A = ∠B

Hence, ∠A =∠B = ∠�

⇒ AB = BC = AC

Therefore, ABC is an equilateral triangle.

24. In an ∆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 that ∆ACE is isosceles.

Solution

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DA || CE [Given]

⇒ ∠1 = ∠4 ..... (i) [Corresponding angles]

∠2 = ∠3...... (ii) [Alternate angles]

But ∠1 = ∠2..... (iii) [AD is the bisector of ∠A]

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

∠3 = ∠4

⇒AC = AE

⇒ ∆ACE is an isosceles triangle.

25. In triangle ABC, bisector of angle BAC meets opposite side BC at

point D. If BD = CD, prove that ∆ABC is isosceles.

Solution

Produce AD upto E such that AD = DE.

In ∆ ABD and ∆ EDC,

AD = DE [ by construction]

BD = CD [given]

∠1 = ∠2 [vertically opposite angles]

∴ ∆ ABD ≅ ∆ EDC [SAS]

⇒ AB = CE...... (i)

And ∠BAD = ∠CAD

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

∴ ∠CED = ∠CAD

⇒ AC = CE...... (ii)

From (i) and (ii)

AB = AC

Hence, ABC is an isosceles triangle.

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26. In ∆ABC, D is point on BC such that AB = AD = BD = DC. Show

that:

∠ADC: ∠C = 4: 1.

Solution

Since AB = AD = BD

∴ ∆ABD is an equilateral triangle.

∴ ∠ADB = 60°

⇒ ∠ADC = 180° - ∠ADB

= 180° - 60°

= 120°

Again in ∆ ADC,

AD = DC

∴ ∠1 = ∠2

But, ∠1 + ∠2 + ∠ADC = 180°

⇒ 2∠1 + 120° = 180°

⇒ 2∠1 = 60°

⇒ ∠1 = 30°

∴ ∠ADC: ∠C = 120° : 30°

⇒ ∠ADC: ∠C = 4 :1

27. Using the information given in each of the following figures, find the

values of a and b.

[Given: CE = AC]

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Solution

(i) In ∆CAE, ∠CAE = ∠AEC = &N5°OQN°

' = 56° [∵ CE = AC]

In ∠BEA, a = 180° - 56° = 124°

In ∆ABE, ∠ABE = 180° - (a + ∠BAE)

= 180° - (124° + 14° )

= 180°- 138° = 42°

(ii)

In ∆ AEB and ∆CAD,

∠EAB = ∠CAD [given]

∠ADC = ∠AEB [∵ ∠ADE = ∠AED {AE=AD} 180° -∠ADE = 180° -

∠AED ∠ADC = ∠AEB]

AE = AD [given]

∴ ∆AEB ≅ ∆CAD [ASA]

AC = AB [By C.P.C.T]

2a + 2 = 7b – 1

⇒ 2a – 7b = - 3..... (i)

CD = EB

⇒ a = 3b..... (ii)

Solving (i) and (ii) we get

a = 9, b = 3

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