NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Triangles |

Exercise |
Ex 5.3 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3

**Question 1.**

**∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that**

**(i) ∆ABD = ∆ACD**

**(ii) ∆ABP = ∆ACP**

**(iii) AP bisects ∠A as well as ∠D**

**(iv) AP is the perpendicular bisector of BC.**

**Solution:
**

**Given**∆ABC and ∆DBC are two isosceles triangles having common

base BC, suchthat AB=AC and DB=OC.

**To prove:**

(i) ∆ABD = ∆ACD

(ii) ∆ABP = ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

**Question 2.**

**AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that**

**(i) AD bisects BC**

**(ii) AD bisects ∠A**

**Solution:**

In ∆ ABD and ∆ ACD, we have

AB = AC (Given)

∠ADB = ∠ADC = 90° (∵ Given AD ⊥BC)

AD = AD (Common)

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

BD=DC (By CPCT)

⇒ AD bisects BC.

∠ BAD = ∠ CAD (By CPCT)

∴ AD bisects ∠A .

**Question 3.**

**Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that**

**(i) ∆ABC ≅ ∆PQR**

**(ii) ∆ABM ≅ ∆PQN**

**Solution:**

**Question 4.**

**BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.**

**Solution:**

In ∆BEC and ∆CFB, we have

**Question 5.**

**ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.**

**Solution:**

In ∆ABP and ∆ACP, We have

AB = AC (Given)

AP = AP (Common)

and ∠APB = ∠APC = 90° (∵ AP ⊥ BC)

∴ ∆ABP ≅ ∆ACP (By RHS congruence axiom)

⇒ ∠B = ∠C (By CPCT)

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