## 21429[EMHL] Re: Three collinear centers

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• Jan 17, 2013
Dear Antreas

--- In Hyacinthos@yahoogroups.com, "Antreas" wrote:
>
> Dear Alex
>
> Angel has tested the loci problems and it seems that
> for all P's the lines are concurrent for the cevian case,
> and for the pedal case as well except for the points P on the
> bisectors and on the circumcircle.
>
> Hmmmmmmm...... Maybe it is true in general ie for any
> circle intersecting the sidelines of the triangle!!
>

A check with GeoGebra:

Let ABC be a triangle and P a point.

Let (Q) be the CEVIAN circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.

For all P (except for the points P on the bisectors) the lines TabTac, TbcTba, TcaTcb are concurrent.

http://amontes.webs.ull.es/geogebra/Hyacinthos21421.html

----------------------

Let (Q) be the PEDAL circle of P and (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the points Tbc, Tba and Tca, Tcb.

For all P (except for the points P on the
bisectors and on the circumcircle) the lines TabTac, TbcTba, TcaTcb are concurrent.

http://amontes.webs.ull.es/geogebra/Hyacinthos21421Pedal.html

------------------------------------------------

Let (Q) be the circle Intersecting the sidelines of the triangle in D, E and F. (Qab), (Qac) the circles touching AB,AC and (Q) internally and let (Tab), (Tac) be the points of contact. Similarly we define the
points Tbc, Tba and Tca, Tcb.

The lines TabTac, TbcTba, TcaTcb are concurrent.

http://amontes.webs.ull.es/geogebra/Hyacinthos21428.html

Best regards,
Angel M.
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