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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.

alexb
Charter Member
1408 posts
Jan-07-05, 08:39 AM (EST)

1. "RE: Equilateral Triangle"
In response to message #0

 Even simpler. Very good.

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.

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 veryfirst version without rereading it. I even forgot to sign it with my name: Narcyz R. Babnis.

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