Prove that it's a tangent of this circle

Hi http://www.mymathforum.com/images/sm...icon_smile.gif

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 http://www.mymathforum.com/images/sm...icon_smile.gif

Peter

PeterWhite

It is Very easy : The angle PAC=25 and it is subtended at the circumference and its sides define the arc PC and the chord PC.This angle ,according to a well known theorem of Euclidean Geometry is equal to the angle formed by the chord PC and the Tangent to the circle at C. since the angle DCP is 25 degrees it follows that the line PC is tangent to the circle at C.
As simple as such.

MINOAS

arc PC =2*25=50
arcAP=90-50=40
angle PBA=40