December 5th, 2017, 10:09 PM  #1 
Newbie Joined: Dec 2017 From: Iran Posts: 3 Thanks: 0  geometry proof triangle
It's an equilateral triangle named "ABC" The center point of the triangle is named "O" The circle that crosses "A" point and "O" crosses "AB" in "M" point and "AC" in "N" point . Prove that AN = BM. Last edited by skipjack; December 6th, 2017 at 05:43 PM. 
December 6th, 2017, 01:24 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,761 Thanks: 696 
Equilateral triangle is congruent to itself by flipping along OA line.

December 6th, 2017, 06:56 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,636 Thanks: 2081 
OA = OB (given) ON = OM (because $\small\angle$NAO = $\small\angle$MAO = 30$^\circ$) $\small\angle$AON = $\small\angle$BOM (because $\small\angle$MON = $\small\angle$BOA = 120$^\circ$) Hence triangles ANO, BMO are congruent (SAS), which implies AN = BM. 
December 24th, 2017, 10:54 AM  #4  
Member Joined: Nov 2016 From: USA Posts: 36 Thanks: 1 
I'm not the OP; however, I'm interested in learning the solution method above. I solved it slightly differently. My diagram is attached. I solved it: angle MBO = angle NAO = 30 ON = OM OA=OB, which means angle BMO = angle ANO Triangle ANO is congruent to Triangle BMO by AAS AN=BM CPCTC Quote:
I understand how angle BOA = 120, but how is it known that angle MON = 120 degrees? Last edited by Seventy7; December 24th, 2017 at 10:56 AM.  
December 24th, 2017, 04:04 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,636 Thanks: 2081 
As triangle ABC is equilateral, $\small{\angle\hspace{1px}}$MAN = 60$^\circ\!$. Opposite angles of the cyclic quadrilateral AMON are supplementary, so $\small{\angle\hspace{1px}}$MON = 120$^\circ\!$. Weren't you using similar reasoning to show that angle BMO = angle ANO? 
December 24th, 2017, 05:53 PM  #6  
Member Joined: Nov 2016 From: USA Posts: 36 Thanks: 1  Quote:
Thank you for this explanation. I "see" it now. Happy Holidays and Happy New Year! Last edited by Seventy7; December 24th, 2017 at 06:07 PM.  
December 24th, 2017, 06:37 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,636 Thanks: 2081  
December 24th, 2017, 11:47 PM  #8  
Member Joined: Nov 2016 From: USA Posts: 36 Thanks: 1  Quote:
Your explanation and reasoning are correct (use opposite supplementary angles of the cyclic quadrilateral AMON; SAS congruency). Thanks again!  
December 26th, 2017, 11:57 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,636 Thanks: 2081 
The AAS approach is okay if it is stated that angle BMO = angle ANO because AMON is a cyclic quadrilateral instead of because OA = OB.


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