Re: [obm-l] Quadrilatero Circunscritivel

[Claudio Buffara <claudio.buffara@xxxxxxxxxxxx>]

on 13.12.04 15:27, Luís Lopes at qed_texte@hotmail.com wrote:

> 
>>> Circle with center in point H is inscribed into convex quadrilateral
>>> ABCD, point H doesn't lie on line AC. Diagonals AC and BD intersect
>>> at point F. Line passing through point F and perpendicular to line
>>> BD, cuts lines AH and CH in points R and S respectively. Prove that
>>> RF=FS.
>>> 
>>> Michel Swift
>> 
> 
> One can view vertices of ABCD quadrilateral as foci and points of
> a rectangular hyperbola  ( B,D  - foci,   A,C - points on
> one branch,   because  AB+CD = BC+AD  or AB-AD = BC-CD = 2a )
Bela sacada!

> Then lines AH, CH are tangents to hyperbola at A and C.
Por que isso eh verdade?
 
> A line perpendicular to BD (main axis) at point F cuts tangents at
> symmetric points wrt BD.
Isso tambem nao eh muito obvio, mas com algum braco deve dar pra demonstrar
por g. analitica. Voce tem uma demonstracao sintetica?
 
 
[]s,
Claudio.


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