Discuss the limit state: 1 When point Q coincides with point C, CP is the tangent of circle 0. At this time, the right movement of point P will

Does not intersect with circle 0. It is easy to get ∠ODQ=∠ODC (point Q coincides with point C).

∠OPC=∠OCD (at this time, the triangle OCP is a right triangle with CH⊥OP).

So ∠OPC=∠ODQ

2. When point P coincides with point B, point P moving to the left will have no intersection with circle 0.

It is easy to get ∠OPC=∠ODQ from the congruence of triangles (at this time, point Q and point P coincide with point B).

3. General state: OQ is connected with ∠QEP=∠DEA (diagonal).

∠COQ=2*∠CDQ (the central angle is twice that of the same arc)

So ∠OCQ=( 180-∠COQ)/2 (triangle OCQ is isosceles)

∠OCQ =( 180-2 *∠CDQ)/2 = 90-∠CDQ

In the triangle DHE, AED=90-∠CDQ.

So ∠OCQ=∠AED=∠PEQ.

So the triangle OCP is similar to the triangle QEP.

So ∠PQE=∠COP=∠DOP.

So the triangle ode is similar to the triangle QPE.

∠OPC and ∠ODQ