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December 5th, 2017, 10:09 PM   #1
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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.
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December 6th, 2017, 01:24 PM   #2
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Equilateral triangle is congruent to itself by flipping along OA line.
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December 6th, 2017, 06:56 PM   #3
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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.
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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 View Post
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
File Type: jpg Geo Eq Triangle.jpg (7.8 KB, 8 views)

Last edited by Seventy7; December 24th, 2017 at 10:56 AM.
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December 24th, 2017, 04:04 PM   #5
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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?
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December 24th, 2017, 05:53 PM   #6
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Quote:
Originally Posted by skipjack View Post
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.

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Originally Posted by skipjack View Post
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.
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December 24th, 2017, 06:37 PM   #7
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Quote:
Originally Posted by Seventy7 View Post
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?
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December 24th, 2017, 11:47 PM   #8
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Quote:
Originally Posted by skipjack View Post
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!
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December 26th, 2017, 11:57 AM   #9
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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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