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• ... BC. ... Dear dam_xoan90 and Francois. A simple elementary proof of this problem, has already been post in Mathlinks Forum, one year ago. Please see at:
Message 1 of 4 , Sep 1, 2007
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--- In Hyacinthos@yahoogroups.com, "dam_xoan90" <dam_xoan90@...> wrote:
>
> Let ABC be an acute triangle and M be the middle point of the side
BC.
> Let H the orthocenter of this triangle and let E the intersection of
> HM with the angle bisector of angle A. Denote by F the orthogonal
> projection of E on AC, prove that HF is perpendicular to AE
>

Dear "dam_xoan90" and Francois.

A simple elementary proof of this problem, has already been post in

Best regards.
Kostas Vittas.
• ... isosceles in A. ... circumcenter. ... side AC and ... uniform ... as choo-c ... correspondence ... and B, so ... AB and ... (B , C ) ... Is it known that
Message 2 of 4 , Sep 2, 2007
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--- In Hyacinthos@yahoogroups.com, "Francois Rideau"
<francois.rideau@...> wrote:
>
> Dear friend
>
> As I am in my dotagen I see choo-choo everywhere!
>
> First, your problem only makes sense when triangle ABC is
isosceles in A.
>
> So in the sequel, I suppose that AB <> AC.
>
> Line HM is on the point A '' symmetric of A wrt the ABC-
circumcenter.
>
> Now I give you a solution based on choo-choo theory.
>
> Given any point P on line HM , we project him orthogonally at Q on
side AC and
> on R on side AB.
>
> When P moves on line HM (as a choo-choo that is to say with a
uniform
> motion!), points Q moves on line AC and R moves on line AB, (also
as choo-c
> hoos!). Then the correspondence g: Q --> R is an affine
correspondence
> between lines AC and AB, that is to say preserves affine ratio.
>
> Look now at special pairs of homologous points in this affine
> correspondence.
>
>
>
> 1° When P is on A'', its projections on line AC and AB are just C
and B, so
> the pair (C,B) is a pair of homologous points.
>
> 2) When P is on H, its projections are B' on line AC and C' on line
AB and
> remind lines BB' and CC' are altitudes of triangle ABC. So the pair
(B', C')
> is another pair of homologous points in this affine correspondence.
>
>
>
Is it known that the envelop of line (QR) when P moves on line (HM)
is a parabol
If P = k H + (1-k) M
Q = (1+k) S_c A +(b²+S_a - k S_c) C
and R = (1+k)S_b A + (S²+S_a - k S_b) B
it is easy to calculate equation of line (QR)
How can we find equation of the envelop?

Michel
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