I've got this task: P is a point in the interior of a square ABCD, such that ∠DCP=∠CAP=25°. What is ∠PBA?
My way of solving is: Consider the circumcircle of CAP. CD is tangential of the circumcircle. Hence, the circumcenter lies on BC, which is perpendicular to CD at D. Also, the circumcenter lies on the perpendicular bisector of AC, which is the line BD. Thus, B is the circumcenter of APC. This shows that BA=BP=BC, so BAP is an isosceles triangle, which gives that ∠BPA=∠BAP=70°. So ∠PBA hast to be 180°-70°-70°=40°.
Now I've got one PROBLEM: How can I prove that it really is the tangent of the circumcircle?
Hope for a good answer