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NCERT Solutions For Class 9 Maths Chapter 5 Triangles Ex 5.3

Here, Below you all know about NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 Question Answer. I know many of you confuse about finding this Chapter 5 Triangles Class Ex 5.3 Of 9 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 9
SubjectMaths
ChapterChapter 5
ExerciseClass 9 Chapter 5 Triangles Exercise 5.3
Number of Questions Solved5
NCERT Solutions For Class 9 Maths Chapter 5 Triangles Ex 5.3

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

(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)

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

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