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February 19th, 2013, 07:54 AM  #1 
Newbie Joined: Feb 2013 Posts: 2 Thanks: 0  Prove that it's a tangent of this circle
Hi 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 C. 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 has 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 Peter 
February 19th, 2013, 08:26 AM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Prove that it's a tangent of this circle
That is true if and only if B is the center of the circle. Is that true?

February 19th, 2013, 08:31 AM  #3 
Newbie Joined: Feb 2013 Posts: 2 Thanks: 0  Re: Prove that it's a tangent of this circle
I think you mean B as the center of the circumcircle is the same as B as the edge of the square?... that's not necessarily the case

February 19th, 2013, 03:14 PM  #4  
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038  Quote:
?DAP = ?PCA = 20°, so DA is a tangent to the circumcircle (by the converse of the alternate segment theorem). Hence B is the center of the circumcircle and so ?PBA = 2*20° = 40°.  
February 21st, 2013, 11:28 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
You reported your own thread as not making sense, but it makes good sense to me (and so I'll leave it).


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