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August 5th, 2012, 03:48 PM   #1
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prove that ( BIOC) cyclic quadrilaterals

prove that ( BIOC) cyclic quadrilaterals
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 August 6th, 2012, 09:54 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: prove that ( BIOC) cyclic quadrilaterals Dunno, but I can't "see or understand" your problem; I guess others can't either....since you got no replies... SO....can you clarify
August 6th, 2012, 01:17 PM   #3
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Re: prove that ( BIOC) cyclic quadrilaterals

According to your information, the drawing should be something like this, it looks very strange... did I draw something wrong? Please Clarify if I made a mistake so I could try to fix it
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August 6th, 2012, 02:53 PM   #4
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Re: prove that ( BIOC) cyclic quadrilaterals

prove that ( HDBF) cyclic quadrilaterals
ANGLE HFG =60
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August 6th, 2012, 02:57 PM   #5
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Re: prove that ( BIOC) cyclic quadrilaterals

Quote:
 Originally Posted by wuzhe According to your information, the drawing should be something like this, it looks very strange... did I draw something wrong? Please Clarify if I made a mistake so I could try to fix it
THAT IS correct

 August 6th, 2012, 06:24 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: prove that ( BIOC) cyclic quadrilaterals Angle at HFG = 60?! So an equilateral triangle? I don't need this headache...all yours Mark!
 August 6th, 2012, 06:36 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: prove that ( BIOC) cyclic quadrilaterals I looked at this some last night, and did consider the equilateral case, which in fact gives a "degenerate" quadrilateral (a triangle) as the inscribed circle and circumscribed circles will both be centered at the in-center of the triangle. I briefly considered a coordinate geometry method, but decided against it as surely there is a much slicker, more elegant way to go here...which of course made me think of "someone else." I could haul out my copy of Euclid's The Elements, but I am lazy today!
 August 6th, 2012, 09:41 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 $^{^{\compose{\rule[-8]{8}{1}}{{\line(8,}}}}" /> BIC = 180° - ($^{^{\compose{\rule[-8]{8}{1}}{{\line(8,}}}}" /> ABC + $^{^{\compose{\rule[-8]{8}{1}}{{\line(8,}}}}" /> ACB)/2 = 180° - (180° - 60°)/2 = 120° and $^{^{\compose{\rule[-8]{8}{1}}{{\line(8,}}}}" /> BOC = 2$^{^{\compose{\rule[-8]{8}{1}}{{\line(8,}}}}" /> BAC = 120°. Hence BIOC is a cyclic quadrilateral.
August 7th, 2012, 05:26 AM   #9
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Re: prove that ( BIOC) cyclic quadrilaterals

Quote:
 Originally Posted by Denis Angle at HFG = 60?! So an equilateral triangle? I don't need this headache...all yours Mark!
F=60
H DOES NOT Equal 60

 August 7th, 2012, 05:50 AM #10 Member   Joined: Jul 2012 Posts: 41 Thanks: 0 Re: prove that ( BIOC) cyclic quadrilaterals I need more Explain skipjack please

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