You are given a circle of center O and a inscribed trapezoid ABCD(AB//CD) with AB<CD. If P is a point on the arc CD which don't belong to A and B and $P_1,P_2,P_3,P_4$ the image of P on the lines AD,BC,BD and AC correspondingly and If AB=a, CD=b and the distance between parallel chords are h, find all the points on the axe of symmetry of the trapezoid ABCD which see the non parallel sides with a right angle (in other words that the angle which is formed when connecting the two endpoints of the non parallel sides, to the point, the resulting angle is equal to 90 degrees) and calculate there distance from AB and CD with a,b, h.
I have thus far managed to prove for this question that $P_1,P_2,P_3, P_4$ are concyclic, as $P_1,P_3,M$ are colinear and $P_2,P_4,M$ are colinear and hence from power of point to circle, we have that they are concyclic. Moreover, I have also proven that the circumscribed circles of triangles $PP_1D$ and $PP_2C$ coincide at point N, as $P_1DNP$ is inscribable and hence $P_1,D,N,P$ are concyclic. Same applies for $P_2,C,N,P$. And hence what I have just stated, holds true. However, I did not manage to solve the question. Could you please help me solve it?