How Can You Prove <)APB+<)CPD=180' in a Base-Quadrilateral Pyramid?

AI Thread Summary
To prove that angle APB plus angle CPD equals 180 degrees in a base-quadrilateral pyramid ABCDS, one must consider the properties of the inscribed sphere tangent at point P on the quadrilateral base ABCD. The relationship between the angles can be derived from the geometric properties of cyclic quadrilaterals and the tangential nature of the sphere. Participants in the discussion express a lack of enthusiasm for geometry and suggest seeking help on specialized mathematics forums. The conversation hints at a connection between the problem and competitive mathematics, particularly in Olympiad contexts. Understanding the geometric relationships and properties is crucial for constructing a valid proof.
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There is a base-quadrilateraled pyramid ABCDS , which base is quadrilateral ABCD. Inscribed sphere is tangent in point P on the ABCD wall. Proof that

<)APB+<)CPD=180'


I don't like geometry and i really don't know how to start.
 
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I guess you are participating both Physics and Mathematics Polish Olympiad... Try to ask this question on the website www.matematyka.org
 
no I'm not
 
That's interesting, because only about 2-3 of your threads are not connected to problems given to solve on the first levels of those olympiads...
 

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