NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Triangles |

Exercise |
Ex 5.2 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.2

**Question 1.**

**In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that**

**(i) OB = OC**

**(ii) AO bisects ∠A**

**Solution:**

(i) In ∆ ABC, we have

AB = AC (Given)

⇒ ∠B = ∠C

(∵ Angles opposite to equal sides are equal)

**Question 2.**

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

**Solution:**

In ∆ABD and ∆ACD, we have ,

DB = DC

∠ ADB = ∠ ADC (∵ D bisect SC)

and AD = AD (Common)

∴ ∆ ABD ≅ ∆ACD (By SAS congruence axiom)

⇒ AB = AC (By CPCT)

Renee,∆ ABC is an isosceles triangle.

**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:**

In ∆ ABE and ∆ ACF, we have

∠ AEB = ∠ AFC (BE ⊥ AC, CF ⊥ AS, each 90°)

∠ A = ∠ A (Common)

and AS = AC (Given)

∴ ∆ABE ≅ ∆ACF (By AAS congruence axiom)

⇒ BE = CF (By CPCT)

**Question 4.** .

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

**(i) ∆ABE = ∆ACF**

**(ii) AB = AC i.e., ABC is an isosceles triangle.**

**Solution:**

In ∆ABE and ∆ACF, we have

∠ AEB = ∠ AFC (Each 90°)

∠ BAE = ∠ CAF (Common)

and BE =CF (Given)

∴ ∆ABE ≅ ∆ACF (By AAS congruence axiom)

∴ AB = AC

So, ∆ABC is isosceles.

**Question 5.**

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

**Solution:**

In ∆ABC, we have

AB=AC (∵ AABC is an isosceles triangle)

∴ ∠ ABC = ∠ ACB …(i)

(∵ Angles opposite to equal sides are equal)

In ∆ DBC, we have

BD = CD (∵ ADBC is an isosceles triangle)

∴ ∠ DBC = ∠ DCB …(ii)

(∵ Angles opposite to equal sides are equal)

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

∠ ABC + ∠ DBC = ∠ ACB + ∠ DCS

⇒ ∠ ABD = ∠ ACD

**Question 6.**

**∆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:**

**Question 7.**

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

**Solution:**

We have, ∠A = 90° (Given)

AB = AC (Given)

⇒ ∠B = ∠C

(∵ Angles opposite to equal sides are equal)

Now, we have

∠A+∠B+∠C = 180° (By ∆ property)

90° + ∠B+ ∠B = 180°

⇒ 2 ∠B = 90°

⇒ ∠B = 45°

∴ ∠B = ∠C = 45°

**Question 8.**

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

**Solution:**

Let ∆ ABC be an equilateral triangle, such that

AB = BC = CA (By property)

Now, AB = AC

⇒ ∠B = ∠C …..(i)

(∵Angles opposite to equal sides are equal)

Similarly, CB = CA

⇒∠A = ∠B …(ii)

(∵ Angles opposite to equal sides are equal)

From Eqs. (i) and (ii), we have

∠A=∠B=∠C …(iii)

Now, ∠A+ ∠B+ ∠C = 180° (By ∆ property)

∠A + ∠A + ∠A = 180° [From Eq. (iii)]

3 ∠A = 180°

∠A = 60°

Hence, ∠ A = ∠B = ∠C = 60°

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