Here, in this circle, AO = OB = CO = OD = radius
From question,
CD = CO = OD
∴ ΔOCD is an equilateral triangle.
∴ ∠OCD = ∠ODC = ∠COD = 60°
Now, consider the cyclic quadrileteral ABCD,
∠CAB + ∠BDC = ∠ACD + ∠ABC = 180° ...(i)
Let ∠ACO = x and ∠BDO = y.
then, in ∠AOC, CO = AO = radius
∴ ∠OAC = ∠ACO = x
Similarly in ΔBOD, ∠ODB = ∠BDO = y
Putting these values in equation (i)
∠CAB + ∠BDC = 180°
∠CAB + ∠BDO + ∠ODC = 180°
x + y + 60° = 180°
∴ x + y = 120° ...(ii)
Now, in ΔCPD,
∠PCD = 180° – ∠ACD = 180° – (x + 60°)
and ∠PDC = 180° – ∠BDC = 180° – (y + 60°)
Sum of angles in ΔCPD = 180°
∴ ∠PCD + ∠CPD + ∠PDC = 180°
180° – (x + 60°) + ∠CPD + 180° – (y + 60°) = 180°
60° – (x + y) + ∠CPD = 0
60° – 120° + ∠CPD = 0
∴ ∠CPD = 60°
Hence, ∠APB = 60°