In ΔACD,
AB = AC [∵ Given]
∠ACD = ∠D …(1) [∵ Angles opposite to equal sides are equal]
In ΔABC,
AB = AC [∵ Given]
∠B = ∠ACB …(2) [∵ Angles opposite to equal sides are equal]
In ΔDBC,
∠D + ∠B + ∠BCD = 180°
⇒ ∠ACD + ∠ACB + BCD = 180° [∵ From the equation (1) and (2)]
⇒ ∠BCD + ∠BCD = 180° [∵ ∠ACD + ∠ACB = ∠BCD]
⇒ 2∠BCD = 180°
⇒ ∠BCD = (180°/2) = 90°
Hence, ∠BCD is a is a right angle.

In ΔABD and ΔACD,
AB = AC [∵ Given]
BD = CD [∵ Given]
AD = AD [∵ Common]
Hence, ΔABD ≅ ΔACD [∵ SSS Congruency Rule]

(ii) In ΔABD ≅ ΔACD, [∵ Proved above]
∠BAD = ∠CAD [∵ CPCT]
In ΔABP and ΔACP,
AB = AC [∵ Given]
∠BAP = ∠CAP [∵ Proved above]
AP = AP [∵ Common]
Hence, ΔABP ≅ ΔACP [∵ SAS Congruency Rule]

(III) In ΔABD ≅ ΔACD [∵ Proved above]
∠BAD = ∠CAD [∵ CPCT]
∠BDA = ∠CDA [∵ CPCT]
Hence, AP bisects both the angles A and D.

(iv) In ΔABP ≅ ΔACP [∵ Proved above]
BP = CP [∵ CPCT]
∠BPA = ∠CPA [∵ CPCT]
∠BPA + ∠CPA = 180° [∵ Linear pair]
⇒∠CPA + ∠CPA = 180° [∵∠BPA = ∠CPA]
⇒ 2∠CPA = 180° ⇒∠CPA = (180°/2 = 90°
⇒ AP is perpendicular to BC. ⇒ AP is Perpendicular bisector of BC. [∵ BP = CP]

(i) Given that : BC = QR
⇒ 1/2 BC = 1/2QR ⇒ BM = QN [∵ AM and PM are medians]
In ΔABM and ΔPQN,
AB = PQ [∵ Given]
AM = PN [∵ Given]
BM = QN [∵ Proved above]
Hence, ΔABM ≅ ΔPQN [∵ SSS Congruency Rule]

(ii) In ΔABM ≅ ΔPQN, [∵ Proved above]
∠B = ∠Q [∵ CPCT]
In ΔABC and ΔPQR,
AB = PQ [∵ Given]
∠B = ∠Q [∵ Proved above]
BC = QR [∵ Given]
Hence, ΔABC ≅ ΔPQR [∵ SAS Congruency Rule]