I will just explain the proof.  You must write it up 
in statements the way you were taught.

There are two cases, 
1. when BC is not a diameter, and 
2. when BC is a diameter. 

Case 1:  BC is not a diameter.



Draw radii OC and OB




We use this theorem:

If a central angle and an inscribed angle of a circle subtend the same arc
or equal arcs, then the central angle is twice the inscribed angle.

Central angle COP and inscribed angle CAP subtend the same arc CP; therefore
the central angle COP is twice the inscribed angle CAP.

Similarly, central angle BOP and inscribed angle BAP subtend the same 
arc BP; therefore the central angle BOP is twice the inscribed angle BAP.

Since inscribed angle CAP = inscribed angle BAP, angles COP and BOP are
equal.  Therefore OP bisects angle BOC.  

Triangle BOC is isosceles because its legs are radii.

Therefore OP is perpendicular to BC because the angle bisector of the
vertex angle of an isosceles triangle is perpendicular to its base.

--------------------

Case 2: BC is a diameter:



Angle BAC is a right angle because it is inscribed in a semicircle.
Since AP bisects angle BAC, it is 45°.    

We use the same theorem we used for the other case.

If a central angle and an inscribed angle of a circle subtend the same arc
or equal arcs, then the central angle is twice the inscribed angle.

45° inscribed angle CAP and central angle COP subtend the same arc CP, so 
angle COP is twice 45° or 90°, a right angle, which is the same as saying 
OP is perpendicular to BC.

Edwin