Welcome to the forum.

There is always one circle that goes through the three points of any triangle. A 4th point may or may not be on this circle too. If it is the quadrilateral is called cyclic. ABCD is such a quadrilateral. All cyclic quadrilaterals have the property that opposite angles add up to 180 degrees. I proved this here:

http://www.mathisfunforum.com/viewtopic.php?id=17799 starting at post 4.

Bob

]]>I think there should be some kind of theorem ... maybe the bob-tister-bob theorem of secants.

Maybe I will make a wikipedia page.

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Drew it up in Geogebra, which gave the two bisectors as being perpendicular to each other.

]]>This time I'm more confident about what to do. (But I haven't fully tried it yet. )

Call angle APB = x and ABP = y

So DAB = x + y => DCB = 180 - (x + y) => BCQ = x + y

So get an expression in terms of x and y for BQC.

Let the bisectors cross at R, and let RQ cross BC at S

Get an expression in terms of x and y for RPB and RQB, then RSB.

Hence for PRQ.

EDIT: Thought I'd better check it ..... it does work.

Bob

]]>Hi; I put the image in for you. Also you need to use the math tags for your latex.

]]>I just don't know how to get this one. This problem in one word=aggggggghhhhhh.

would power of a point help?

thanks!