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Subject: "Equilateral Triangle: Another Proof"     Previous Topic | Next Topic
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narbab
guest
Jan-07-05, 08:38 AM (EST)
 
"Equilateral Triangle: Another Proof"
 
   Let:

  • Q lie on AC and N lie outside ABC;

  • AQN be congruent to BPM;

  • D be the centre of AQN.

Then:


  • DF passes through E (the centre of symmetry between triangles AQN and PMB);

  • BCD is congruent to ACF.

So:


  • DCF is equilateral and CE is its height.


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alexb
Charter Member
1408 posts
Jan-07-05, 08:39 AM (EST)
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1. "RE: Equilateral Triangle"
In response to message #0
 
   Even simpler. Very good.


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rewboss
guest
Jan-07-05, 05:37 PM (EST)
 
3. "RE: Equilateral Triangle"
In response to message #0
 
  

  • D be the centre of
    AQN.

I realise this isn't very significant, but in the diagram you have D as the centre of PMB, and F as the centre of AQN.


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narbab
guest
Jan-10-05, 08:16 AM (EST)
 
4. "RE: Equilateral Triangle"
In response to message #3
 
   >I realise this isn't very significant, but in the diagram
>you have D as the centre of PMB, and F as the centre of AQN.

Of course. I was so excited with this idea that I posted the very
first version without rereading it.
I even forgot to sign it with my name: Narcyz R. Babnis.


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