Let P be a point inside triangle ABC, and suppose lines AP, BP, CP meet the circumcircle again at T, S, R (here T=A, S=B, R=C). Let U be any point in the interior of PT. A line through U parallel to AB meets CR at W, and the line through U parallel to AC meets BS again at V. Finally, the line through B parallel to CP and the line through C parallel to BP intersect at point Q. Given that RS and VW are parallel, prove that ∠CAP=∠BAQ. geometrycircumcirclegeometric transformationsimilar trianglesgeometry proposed