My Math Forum Circle, Triangle, Symmetry

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November 17th, 2013, 10:09 AM   #1
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Circle, Triangle, Symmetry

Please solve this problem for me?
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November 17th, 2013, 10:11 AM   #2
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Math Focus: Geometry
Re: Circle, Triangle, Symmetry

Sorry for attachment. Here smaller one.
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 November 27th, 2013, 05:16 PM #3 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry Re: Circle, Triangle, Symmetry I solved this problem. I hope some one will solve this --> viewtopic.php?f=13&t=44386
 November 29th, 2013, 11:29 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 Can you post your proof?
December 2nd, 2013, 01:49 PM   #5
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Re: Circle, Triangle, Symmetry

Sure.
[attachment=0:1np88ziz]triangleproblem222.png[/attachment:1np88ziz]
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 December 3rd, 2013, 12:43 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 What does "From HCID'" mean in the last line of your proof?
 December 5th, 2013, 07:42 AM #7 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry Re: Circle, Triangle, Symmetry The sum of angles CID' + ID'H + D'HC + HCI = 360 and we get alpha + beta + gamma + theta = 180 degrees. Therefore D'_1DD'_2 is triangle.
 December 5th, 2013, 10:05 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 You hadn't found angles CID' and D'HC. What values did you use for them?
 December 5th, 2013, 02:36 PM #9 Senior Member     Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry Re: Circle, Triangle, Symmetry angles CID' and D'HC are 90-theta and 90-gamma resp.
 December 5th, 2013, 03:12 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 But as you seemed to find that HCI = 180 - (? + ?), and that ID'H = ? + ? + ? + ?, adding up the angles of HCID' then gives (90 - ?) + (? + ? + ? + ?) + (90 - ?) + (180 - (? + ?)), which is 360 anyway, without implying that ? + ? + ? + ? = 180.

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