Re: [obm-l] Quadrilatero Circunscritivel

[Luís Lopes <qed_texte@xxxxxxxxxxx>]

Sauda,c~oes,

Oi Claudio,

>Isso tambem nao eh muito obvio, mas com algum braco deve dar pra demonstrar
>por g. analitica. Voce tem uma demonstracao sintetica?

Não.

[]'s
Luis

>From: Claudio Buffara <claudio.buffara@terra.com.br>
>Reply-To: obm-l@mat.puc-rio.br
>To: <obm-l@mat.puc-rio.br>
>Subject: Re: [obm-l] Quadrilatero Circunscritivel
>Date: Mon, 13 Dec 2004 17:32:24 -0200
>
>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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