Cyclic quad. problem!!!!!!!!!!

Join AD, OC, BC

For quad ADPQ,
angle PQA = 90 (given)
angle PDA = 90 (angle in semi-circle)
so quad ADPQ is cyclic quad. (opp. angles are supp.)

Let angle PAD = x,
then angle PQD = x (angle in same segment) . . . (i)

For quad BCPQ,
angle BQP = 90 (given)
angle BCP = 90 (angle in semi-circle)
so quad BCPQ is also cyclic quad. (opp. angles are supp.)

angle PQC
= angle PBC (angle in same segment, for quad BCPQ)
= angle PAD (angle in same segment for quad ABCD)
= angle PQD (proved in (i))
therefore, PQ bisects angle DQC

angle COD
= 2 * angle CAD . . . . . . . . . (angle at center = twice angle at circumference)
= 2x . . . . . . . . . . . . . . . . . .(angle CAD = x, assumption)
= angle PQD + angle PQC . .(proved)
= angle CQD
therefore, OQCD are concyclic points. (converse of angle in same segment)