Question

In the given figure, AB = AC, BN and CM are bisectors of ∠ABC and ∠ACB respectively. If AB and AC are the bisectors of ∠MAO and ∠OAN respectively, then which of the following is always correct?

• A. ΔBOC ≅ ΔNOC
• B. ΔBQC ≅ ΔAOB
• C. ΔAMB ≅ ΔANC
• D. ΔAOQ ≅ ΔQAN

Answer 1

The correct option is C ΔAMB ≅ ΔANC

AB = AC (Given)
∴ ∠ABC = ∠ACB (Angles opposite to equal sides are equal) …..(i)
Also, ∠ABN = ∠NBC and ∠MCB = ∠ACM
∴ ∠ABN = ∠NBC = ∠MCB = ∠ACM [From (i)] …..(ii)
In ΔBPC and ΔCQB,
∠BCP = ∠CBQ [From (ii)]
∠PBC = ∠QCB [From (i)]
BC = CB (Common)
∴ ΔBPC ≅ ΔCQB (ASA congruence rule)
⇒ BP = CQ (CPCT)
⇒ AP = AQ (∵ AB = AC) …..(iii)
Now, AB and AC are bisectors of ∠MAO and ∠OAN respectively.
∴ ∠MAP = ∠PAO and ∠NAQ = ∠QAO
In ΔABC, the angle bisectors PC and QB meet OA at O. This implies that OA is also an angle bisector of ∠BAC, i.e., ∠PAQ.
∴ ∠MAP = ∠PAO = ∠NAQ = ∠QAO …..(iv)
In ΔPAO and ΔQAO,
∠PAO = ∠QAO [From (iv)]
AP = AQ [From (iii)]
AO = AO (Common)
∴ ΔPAO ≅ ΔQAO (SAS congruence rule)
⇒ ∠APO = ∠AQO (CPCT)
⇒ 180° – ∠APO = 180° – ∠AQO
⇒ ∠APM = ∠AQN ….(v)
In ΔPAM and ΔQAN,
∠MAP = ∠NAQ [From (iv)]
AP = AQ [From (iii)]
∠APM = ∠AQN [From (v)]
∴ ΔPAM ≅ ΔQAN (ASA congruence rule)
⇒ AM = AN (CPCT) …..(vi)
In ΔAMB and ΔANC,
∠MAB = ∠NAC [From (iv)]
AM = AN [From (vi)]
AB = AC (Given)
∴ ΔAMB ≅ ΔANC (SAS congruence rule)
Hence, the correct answer is option (c).