Let k and q be symmetrical points of the median line of points A and P about BC,

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