RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2 (Updated for 2021-22)

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2: We have the perfect Maths guide for all the Class 9 students, i.e., RD Sharma Solutions Class 9 Maths. You can practice the questions and clear all your doubts. RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2 solutions are desiged as per the current CBSE Syllabus by our subject matter experts. To know more, read the whole blog. 

Download RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2 PDF

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2

Access answers of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2

Question 1.
BD and CE are bisectors of ∠B and ∠C of an isosceles ∠ABC with AB = AC. Prove that BD = CE.
Solution:
Given : In ∆ABC, AB = AC
BD and CE are the bisectors of ∠B and ∠C respectively
To prove : BD = CE
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
∴ \(\frac { 1 }{ 2 }\) ∠B = \(\frac { 1 }{ 2 }\) ∠C
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 1
∠DBC = ∠ECB
Now, in ∆DBC and ∆EBC,
BC = BC (Common)
∠C = ∠B (Equal angles)
∠DBC = ∠ECB (Proved)
∴ ∆DBC ≅ ∆EBC (ASA axiom)
∴ BD = CE

Question 2.
In the figure, it is given that RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that: ∆RBT = ∆SAT.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 2
Solution:
Given : In the figure, RT = TS
∠1 = 2∠2 and ∠4 = 2∠3
To prove : ∆RBT ≅ ∆SAT
Proof : ∵ ∠1 = ∠4 (Vertically opposite angles)
But ∠1 = 2∠2 and 4 = 2∠3
∴ 2∠2 = 2∠3 ⇒ ∠2 = ∠3
∵ RT = ST (Given)
∴∠R = ∠S (Angles opposite to equal sides)
∴ ∠R – ∠2 = ∠S – ∠3
⇒ ∠TRB = ∠AST
Now in ∆RBT and ∆SAT
∠TRB = ∠SAT (prove)
RT = ST (Given)
∠T = ∠T (Common)
∴ ∆RBT ≅ ∆SAT (SAS axiom)

Question 3.
Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.
Solution:
Given : Two lines AB and CD intersect each other at O such that AD = BC and AD \(\parallel\)
BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2 – 3
To prove : AB and CD bisect each other
i. e. AO = OB and CO = OD
Proof: In ∆AOD and ∆BOC,
AD = BC (Given)
∠A = ∠B (Alternate angles)
∠D = ∠C (Alternate angles)
∴ ∆AOD ≅ ∆BOC (ASA axiom)
AO = OB and AO = OC (c.p.c.t.)
Hence AB and CD bisect each other.

This is the complete blog on RD Sharma for Class 9 Solutions Chapter 12 Exercise 12.2. To know more about the CBSE Class 9 Maths exam, ask in the comments. 

FAQs on RD Sharma Class 9 Solutions Chapter 12 Exercise 12.2