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 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: 6,820 Thanks: 722 Equilateral triangle is congruent to itself by flipping along OA line.
 December 6th, 2017, 06:56 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 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
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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:
 Originally Posted by skipjack 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.
I am trying to prove this with SAS like the quoted text. I can't see why the sandwiched angle AON = angle BOM?

I understand how angle BOA = 120, but how is it known that angle MON = 120 degrees?
Attached Images
 Geo Eq Triangle.jpg (7.8 KB, 8 views)

Last edited by Seventy7; December 24th, 2017 at 10:56 AM.

 December 24th, 2017, 04:04 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,965 Thanks: 2214 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? Thanks from Seventy7
December 24th, 2017, 05:53 PM   #6
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Quote:
 Originally Posted by skipjack Weren't you using similar reasoning to show that angle BMO = angle ANO?
I used different reasoning. My reasoning for those two angles being congruent: The line segments AO and BO both extend from the equilateral triangle's angles to its center, thus are congruent, which makes congruent the angles BMO and ANO. That gave me two congruent angles and a non-included congruent side on each triangle, so AAS congruency.

Quote:
 Originally Posted by skipjack Opposite angles of the cyclic quadrilateral AMON are supplementary, so $\small{\angle\hspace{1px}}$MON = 120$^\circ\!$.
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
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Quote:
 Originally Posted by Seventy7 The line segments AO and BO both extend from the equilateral triangle's angles to its center, thus are congruent, which makes congruent the angles BMO and ANO.
That's unclear. What theorem are you using to deduce that angles BMO and ANO are congruent?

December 24th, 2017, 11:47 PM   #8
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Quote:
 Originally Posted by skipjack That's unclear. What theorem are you using to deduce that angles BMO and ANO are congruent?
Thank you for pointing this out. My proof above (AAS) is not correct.

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: 20,965 Thanks: 2214 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. Thanks from Seventy7

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