R is the radius of the circumscribed circle of △ABC, then
OA=OB=OC=OK=R。
∵ quadrilateral KQPA is a rectangle, ∴QP=KA,
At △AOK, OA=OK=R,
∠AOK =∠AOB-∠KOB =∠AOB-∠AOC = 2∠BCA-2∠ABC≥60。
∴KA≥R,QP≥R.
According to the trigonometric inequality,
OP+R=OQ+OC >QC=QP+PC≥R+PC,
∴op>; Personal computer
∴∠pco>; ∠COP。
∠cab = 1/2 *∠BOC = 1/2 *( 180-2∠POC)= 90-∠POC,
That is ∠ cab+∠ COP