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# RD Chapter 12- Heron-s Formula Ex-12.2 Interview Questions Answers

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Question 1 : BD and CE are bisectors of ∠B and ∠C of an isosceles ∠ABC with AB = AC. Prove that BD = CE.

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)
∴   ∠B =  ∠C
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)
∴   ∠B =  ∠C
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)
∴ ∠B = ∠C ∠DBC = ∠ECB
Now, in ∆DBC and ∆EBC,
BC = BC (Common)
∠C = ∠B (Equal angles)
∠DBC = ∠ECB (Proved)
∴ ∆DBC ≅ ∆EBC (ASA axiom)
∴ BD = CE

∠DBC = ∠ECB
Now, in ∆DBC and ∆EBC,
BC = BC (Common)
∠C = ∠B (Equal angles)
∠DBC = ∠ECB (Proved)
∴ ∆DBC ≅ ∆EBC (ASA axiom)
∴ BD = CE

∠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. 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.

Given : Two lines AB and CD intersect each other at O such that AD = BC and AD ∥
BC To prove : AB and CD bisect each other
i. e. AO = OB and CO = OD
Proof: In ∆AOD and ∆BOC,
∠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.

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