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February 19th, 2013, 07:54 AM   #1
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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
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February 19th, 2013, 08:26 AM   #2
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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?
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February 19th, 2013, 08:31 AM   #3
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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
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February 19th, 2013, 03:14 PM   #4
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Quote:
Originally Posted by Peter White
Also, the circumcenter lies on the perpendicular bisector of AC, which is the line BD.
That proves B is the circumcenter, but as BD doesn't contain P, the argument below is a bit more convenient.

?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.
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February 21st, 2013, 11:28 AM   #5
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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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