Re: Prependicular lines
- --- In Hyacinthos@yahoogroups.com, "dam_xoan90" <dam_xoan90@...> wrote:
> Let ABC be an acute triangle and M be the middle point of the side
> Let H the orthocenter of this triangle and let E the intersection ofDear "dam_xoan90" and Francois.
> 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
A simple elementary proof of this problem, has already been post in
Mathlinks Forum, one year ago.
Please see at: http://www.mathlinks.ro/Forum/viewtopic.php?t=109749
- --- In Hyacinthos@yahoogroups.com, "Francois Rideau"
>isosceles in A.
> Dear friend
> As I am in my dotagen I see choo-choo everywhere!
> First, your problem only makes sense when triangle ABC is
> So in the sequel, I suppose that AB <> AC.
> Line HM is on the point A '' symmetric of A wrt the ABC-
>side AC and
> 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
> on R on side AB.uniform
> When P moves on line HM (as a choo-choo that is to say with a
> motion!), points Q moves on line AC and R moves on line AB, (alsoas choo-c
> hoos!). Then the correspondence g: Q --> R is an affinecorrespondence
> between lines AC and AB, that is to say preserves affine ratio.and B, so
> Look now at special pairs of homologous points in this affine
> 1° When P is on A'', its projections on line AC and AB are just C
> the pair (C,B) is a pair of homologous points.AB and
> 2) When P is on H, its projections are B' on line AC and C' on line
> 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?