place r′ so that qr′ is parallel to the tangent.
now we get that ∠r′py=∠pr′q, where y is on the tangent, on the opposite side of p from x.
also, because of the symmetry, ∠r′py=∠qpx.
finally, note that no matter where we place r, as long as it's on the correct arc, we have ∠prq=∠pr′q.
stringing all these equalities together, we end up with ∠prq=∠qpx.
this picture will show how its going to look like.
hope this by jack monhen
![Hey me guys in construction with the of compass](/tpl/images/02/03/iCDPh5v3ORju2ZVE.jpg)