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## Fwd: Re: two excircles

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Message 1 of 1 , Jul 2, 2007
>Date: Mon, 02 Jul 2007 08:33:21 -0000
>From: "alexey_zaslavsky" <alexey_zaslavsky@...>
>To: Hyacinthos-owner@yahoogroups.com
>Subject: Re: two excircles
>
>Dear Jean-Pierre and Francois!
>>
>>
>> Here are some remarks (I've changed the notations and I didn't find
>a
>> synthetic proof)
>> Let I,Ia,Ib,Ic be the incenter and excenters; U,U' the feet on BC
>of
>> the internal and external A-bisectors
>> Then there exists
>> - an hyperbola h with center U going through the 4 projections of I
>and
>> Ia upon AB and AC (the asymptots are BC and the other common
>tangent
>> through U to the incircle and the A-excircle)
>> - an hyperbola h' with center U' going through the 4 projections of
>Ib
>> and Ic upon AB and AC (the asymptots are BC and the other common
>> tangent through U' to the B- and C-excircles)
>>
>> Consider a point M lying on h; the parallel and antiparallel to BC
>> through M intersect AB and AC at four points lying on a circle, and
>> this circle touches the incircle and the A-excircle (and the line
>> through the contact points goes through U)
>> Consider a point M' lying on h'; the parallel and antiparallel to
>BC
>> through M' intersect AB and AC at four points lying on a circle,
>and
>> this circle touches the B-excircle and the C-excircle (and the line
>> through the contact points goes through U')
>>
>I can propose a simple construction of circles touching A-excircle
>and B-excircle of the triangle ABC.
>Let P be an arbitrary point on the radical axis of excircles, C_0 ---
>the midpoint of AB. Q is the common point of perpendicular from P to
>AB and CC_0. The line passing through Q and parallel to AB intersect
>AC and BC in points X and Y. Then the circle XPY touches two
>excircles.
>In fact as P in the radical axis there exists an inversion with
>center P conserving both excircles. It isn't difficult to note that
>this inversion transforms the circle XPY to the line AB.
>
>Sincerely Alexey

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