NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2.

- Triangles Class 9 Ex 7.1
- Triangles Class 9 Ex 7.2
- Triangles Class 9 Ex 7.3
- Triangles Class 9 Ex 7.4
- Triangles Class 9 Ex 7.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Triangles |

Exercise |
Ex 7.2 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

Ex 7.2 Class 9 Maths Question 1.

In an Isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that

(i) OB = OC

(ii) AO bisects ∠A

Solution:

**(i)** In ΔABC, we have AB = AC (Given)

=> ∠B = ∠C

\(\frac { 1 }{ 2 } \)∠B = \(\frac { 1 }{ 2 } \)∠C

=> ∠OBC = ∠OCB

and ∠OBA = ∠OCA

(∵ OB and OC are bisectors of ∠B and ∠C respectively).

(∵ ∠OBC = \(\frac { 1 }{ 2 } \)∠B and ∠OCB = \(\frac { 1 }{ 2 } \)∠C)

=> OB – OC (∵ Sides opposite to equal angles are equal)

**(ii)** In Δ ABO and ΔACD, we have

AB = AC (Given)

∠OBA = ∠OCA [From Eq. (i)]

=> OB = OC [From Eq. (ii)]

ΔABO ≅ ΔACO (By SAS congruence axiom)

=> ∠BAO = ∠CAO (By CPCT)

=> AO is the bisector of ∠BAC.

Ex 7.2 Class 9 Maths Question 2.

In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that AABC is an isosceles triangle in which AB = AC.

Solution:

Given : AD is the perpendicular bisector of BC;

To prove : In Δ ABC is an isosceles triangle. i.e.,

AB = AC

Proof : In Δ ADB and Δ ADC,

AD = AD [common]

BD = DC (Given)

and ∠ADB = ∠ADC [each= 90°]

Δ ADB ≅ Δ ADC (By SAS congruence axiom)

=> AB = AC (By CPCT)

So, Δ ABC is an isosceles triangle.

Ex 7.2 Class 9 Maths Question 3.

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively (see figure). Show that these altitudes are equal.

Solution:

Given: ΔABC is an isosceles triangle in which

AB = AC,BE ⊥ AC and CF ⊥ AB

To prove : BE = CF

Proof : In Δ ABE and Δ ACE,

AB – AC given]

∠AEB = ∠AFC [each 90 °, BE ⊥ AC and CF ⊥ AB]

∠BAE = ∠CAF [common angle]

∴ Δ ABE ≅ Δ ACF [by AAS congruence rule]

Then BE = CF [by CPCT] Hence proved.

Ex 7.2 Class 9 Maths Question 4.

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that

(i) ΔABE = ΔACF

(ii) AB = AC

i.e., ΔABC is an isosceles triangle.

Solution:

Given: ΔABC in which BE ⊥ AC and CF⊥ AB, such that BE = CF.

To prove : (i) Δ ABE ≅ Δ ACF

(ii) AB = AC

Proof: (i) In Δ ABC and Δ ACF

∠AEB= ∠AFC [each 90 °]

∠BAE = ∠CAF [common angle]

and BE = CF (Given)

Δ ABE ≅ Δ ACF [by AAS congruence rule]

(ii) From part (i), Δ ABE ≅ Δ ACF

AB = AC [by CPCT]

Ex 7.2 Class 9 Maths Question 5.

ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Solution:

Consider Δ ABC, we have

AB= AC

=> ∠ABC = ∠ACB …(i) [∵ angles opposite to equal sides are equal]

Consider ADBC , we have

BD = CD

=> ∠DBC = ∠DCB …(ii) [∵ angles opposite to equal sides are equal]

Adding eqs. (i) and (ii), we get

∠ABC + ∠DBC = ∠ACB + ∠DCB

=> ∠ABD = ∠ACD

Ex 7.2 Class 9 Maths Question 6.

Δ ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Solution:

In ΔABC, we have AB = AC (Given)

=> ∠ACB = ∠ABC ………..(i) (∵Angles opposite to equal sides are equal)

Now, AB = AD (Given)

∴ AC = AD (∵ AB = AC)

Now, in ∠ADC, we have

AD= AC (from above)

=> ∠ACD = ∠ADC ……..(ii) (∵ Angles opposite to equal sides are equal)

On adding eqs. (i) and (ii), we have

∠ACB + ∠ACD = ∠ABC + ∠ADC

=»∠BCD = ∠ABC + ∠BDC (∵ ∠ADC = BDC)

Adding ∠BCD on both sides, we have

∠BCD + ∠BCD = ∠ABC + ∠BDC + ∠BCD

=> 2∠BCD = 180° (By Δ property)

∠BCD = 90°

Ex 7.2 Class 9 Maths Question 7.

ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Solution:

Given, Δ ABC is a right angled triangle in which ∠A = 90° and AB = AC.

Then, ∠C = ∠B (i) [since, angles opposite to equal sides of a triangle are equal]

Now, ∠A + ∠B + ∠C = 180°[since, sum of three angles of a triangle is 180°]

=> 90 °+∠B + ∠B = 180° [from eq. (i)]

=> 2∠B =90° ∴ ∠B =45°

Hence, ∠B = 45° and ∠C = 45°

Ex 7.2 Class 9 Maths Question 8.

Show that the angles of an equilateral triangle are 60° each.

Solution:

Given: Δ ABC is an equilateral triangle.

To prove : ∠A = ∠B = ∠C = 60°

Proof : Since, all three angles of an equilateral triangle are equal. i.e., ∠A=∠B=∠C …(i)

∴ Z∠A + ∠B + ∠C = 180° [since, sum of three angles of a triangle is 180°]

=> ∠A + ∠A + ∠A = 180° [from eq. (i)]

3∠A = 180°

∠A \(\frac { 180 }{ 3 } \)

=>∠A = 60°

Hence, ∠A = ∠B = ∠C = 60°

**Hence proved**

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