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Re: [EMHL] two triangles

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• Dear Alexey I try to read your proof with some difficulties. 1° You have strange notations for your points. It is usual to label the centroid G, the
Message 1 of 20 , Oct 6, 2007
Dear Alexey
I try to read your proof with some difficulties.
1° You have strange notations for your points.
It is usual to label the centroid G, the circumcenter O, the orthocenter H
and so on..
As for the Lemoine point, one label it K, I don't know why? Maybe L is kept
for the de Longchamps point
Sure there are not enough letters to label all Kimberling little friends!
Besides you label L the Lemoine point in your first post and K in your
proof.
2° I find your triangle A'B'C' a little intricate to define.
It is sufficient to prove your property with any triangle A'B'C' with sides
proportional to the ABC-medians and as you notice it, the pedal triangle of
the ABC-Lemoine point K is more suitable to do the job.

So in the sequel, I label G, O, K the centroid, the circumcenter and the
Lemoine point of triangle ABC.
A'B'C' is the pedal point of K;
So the centroid of A'B'C' is G' = K.
The circumcenter O' of A'B'C' is the middle point of GK.
So if K' is the Lemoine point of A'B'C', your property is equivalent to
prove that KK' is parallel to the Euler line GO of triangle ABC.
Friendly
Francois

[Non-text portions of this message have been removed]
• I correct a little typo ... Of course A B C is the pedal triangle of K. So the centroid of A B C is G = K. ... [Non-text portions of this message have been
Message 2 of 20 , Oct 6, 2007
I correct a little typo

On 10/7/07, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Alexey
> I try to read your proof with some difficulties.
> 1° You have strange notations for your points.
> It is usual to label the centroid G, the circumcenter O, the orthocenter H
> and so on..
> As for the Lemoine point, one label it K, I don't know why? Maybe L is
> kept for the de Longchamps point
> Sure there are not enough letters to label all Kimberling little friends!
> Besides you label L the Lemoine point in your first post and K in your
> proof.
> 2° I find your triangle A'B'C' a little intricate to define.
> It is sufficient to prove your property with any triangle A'B'C' with
> sides proportional to the ABC-medians and as you notice it, the pedal
> triangle of the ABC-Lemoine point K is more suitable to do the job.
>
> So in the sequel, I label G, O, K the centroid, the circumcenter and the
> Lemoine point of triangle ABC.
> A'B'C' is the pedal point of K;

Of course A'B'C' is the pedal triangle of K.

So the centroid of A'B'C' is G' = K.
> The circumcenter O' of A'B'C' is the middle point of GK.
> So if K' is the Lemoine point of A'B'C', your property is equivalent to
> prove that KK' is parallel to the Euler line GO of triangle ABC.
> Friendly
> Francois
>

[Non-text portions of this message have been removed]
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