The square ABCD is inscribed on the circle O, P is on the inferior arc AB, connecting PD and AC to Q, and PQ = OQ Sorry, I can't draw a picture

The square ABCD is inscribed on the circle O, P is on the inferior arc AB, connecting PD and AC to Q, and PQ = OQ Sorry, I can't draw a picture

Let's see if you can understand this solution. The angle OPQ = angle QoP, we also know that the angle OPQ = angle ODP, the angle ODP = 1 / 2 angle POB, the angle POB + the angle QoP = 90 degrees, so we know that the angle QoP = 30 degrees, so the triangle OPQ is two triangles whose base angle is 30 degrees and vertex angle is 120 degrees. If OQ = x, Op = root 3 times of x = od = CD divided by root