Given: ABCD is a parallelogram. The bisectors of and meet at .
To prove: (i) (ii) and (iii)

Proof:

∴ ​∠A = ​∠C and ∠B =​ ∠D … Opposite angles
And ​∠A + ​∠B = 180° … Adjacent angles
∴ ​∠B = 180° − ∠A
180° − 60° = 120° … as ∠A = 60°
∴ ​∠A = ​∠C = 60° and ∠B =​ ∠D​ = 120°
(i) In ∆ APB,

∠​PAB = = 30°and ∠PBA = = 60°
∴​ ∠​APB​ = 180° − (30° + 60°) = 90°
(ii) In ∆ ADP, ∠​PAD = 30° and ∠ADP = 120°
∴ ∠APB = 180° − (30° + 120°) = 30°

Hence,

∠​PAD = ​∠APB = ​30°
Hence, ∆ADP is an isosceles triangle and AD = DP.
In ∆ PBC,

∠​ PBC = 60°

∠​ BPC = 180° − (90° +30°) = 60°and∠​ BCP = 60° …Opposite angle of ∠A
∴ ∠ PBC = ∠​ BPC = ∠​ BCP
Hence, ∆PBC is an equilateral triangle and, therefore, PB = PC = BC.​
(iii) DC = DP + PC
From (ii), we have

DC = AD + BC …AD = BC
DC = AD + AD

DC = 2 AD