Assuming that PQ is not perpendicular to PC, we can draw a straight line PQ'⊥PC, which intersects AB at q'.

Then: q', b, p, c four-point * * * circle,

Inferred from the theorem of circle angle, ⊿P Q'C∽⊿ABD,

According to the nature of similar triangles, pq'/PC = ad/ab,

And because PQ/PC=AD/AB

∴ point q' coincides with point q,

∴∠QPC=90。