Well... let's take three other points on the circumcircle....
The perpendicular bisectors of the sides of ABC intersect
the circumcircle at A',B',C' on the positive sides of BC,CA,AB, resp.
(ie both A,A' are on the same side of BC etc)
The Simson lines of A',B',C' are concurrent at a point P.
Note: If A",B",C" are the other intersections of the perp. bisectors
withthe circumcircle, then the Simson Lines of A",B",C" bound a
triangle which is perspective with the pedal triangle of O (medial
triangle) at P.
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.
The line PPa intersects the circumcircle at A' on the positive side
of BC (ie both A,A' are on the same side of BC). Similarly B',C'.
Which is the locus of P such that the Simson lines of A',B',C' are
On Thu, Mar 7, 2013 at 7:56 AM, rhutson2 <rhutson2@...> wrote:
> They are not perspective. A*B*C* is also the circumcevian triangle of
> X(56). If mixtilinear excircles are used instead, touching the circumcircle
> at A**,B**,C**, then A**B**C** is the circumcevian triangle of X(55). The
> triangle bounded by the Simson lines of A**,B**,C** is also not perspective
> to A'B'C' or A"B"C".
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> > Let ABC be a triangle and A'B'C' the cevian triangle of I.
> > The three mixtilinear incircles touch the circumcircle at A*,B*,C*
> > respective to A,B,C.
> > The Simson lines of A*,B*,C* bound a triangle A"B"C".
> > Are the triangles A'B'C', A"B"C" perspective?
> > APH