- --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>

As Francisco pointed out: "this follows inmediately from Desargues

> Let ABC, A'B'C' be two homothetic triangles

> and X,Y,Z three collinear points.

>

> Are the triangles bounded by the lines

> (AX, BY, CZ) and (A'X, B'Y, C'Z) perspective?

theorem for any ABC and A'B'C'"

We can use it in the reference triangle ABC to get points,

from two given triangles and three collinear points.

Here is an example:

Inside ABC exists a unique point P such that:

Three Congruent (equal) circles (A'),(B'),(C') concur at P,

and circle (A') touches the sides of the angle A,

(B') of angle B and (C') of angle C.

[A', B', C' are the centers of the circles]

Now, let Lp a line passing through P and

intersecting the circles (A'),(B'),(C') at

X,Y,Z resp. (other than P).

The triangles bounded by the lines

(AX, BY, CZ) and (A'X, B'Y, C'Z) are perspective.

Which is the locus of the perspectors

as L moves around P?

APH [APH]:

Let ABC be a triangle and A'B'C' the cevian triangle of I.

Denote:

Bc, Cb = the reflections of B, C in CC', BB', resp.

Oa = the circumcenter of ABcCb. Similarly Ob, Oc.

OaObOc and the excentral triangle IaIbIc are homothetic.

Which point is the homothetic center wrt triangles:

1. ABC (on its OI line)

2. OaObOc (on its Euler line)

3. IaIbIc (on its Euler line) ??

APH[César Lozada]

1. ABC (on its OI line)

X(3576)2. OaObOc (on its Euler line)

X(381)3. IaIbIc (on its Euler line)

X(381)CL