Because if BC is the diameter, then △ABC will be inscribed on the circle O (△ABC is the right triangle and BC is the hypotenuse).
Proof: As shown in Figure AB, it is the tangent of circle O.
∴∠3=∠BCA
∠∠BCA+∠ABC = 90
∴∠3+∠ABC=90
∴AP⊥BC? Q is the midpoint of AB.
∴AQ=QP
∴∠3=∠4
AO = PO
∴∠ 1=∠2? Similarly < 3+<1= 90
∴∠2+∠4=∠ 1+∠3=90
∴QP⊥PO
∴PQ is the tangent of circle O.