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

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

NCERT TEXTBOOK EXERCISES

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

Solution:

In ∆ ABD and ∆ ACD, we have

AB = AC (Given)
∴ ∆ ABD ≅ ∆ ACD (By RHS congruence axiom)
BD=DC (By CPCT)

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)

## NCERT Solutions for Class 9 Maths Chapter 5 Triangles Exercise 5.3 PDF

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