## Τετάρτη, 24 Απριλίου 2013

### EXCENTRAL TRIANGLE

Let ABC be a triangle, IaIbIc the excentral triangle and Q the circumcenter of IaIbIc.

Denote:

Ab, A'b = the intersections of the line IaIc with the excircle (Ic) (Ab near B)

Ac, A'c = the intersections of the line IaIb with the excircle (Ib) (Ac near C)

Bc, B'c = the intersections of the line IbIa with the excircle (Ia) (Bc near C)

Ba, B'a = the intersections of the line IbIc with the excircle (Ic) (Ba near A)

Ca, C'a = the intersections of the line IcIb with the excircle (Ib) (Ca near A)

Cb, C'b = the intersections of the line IcIa with the excircle (Ia) (Cb near B)

A1, A2 = the circumcenters of IaAbAc, IaA'BA'c, resp.

B1, B2 = the circumcenters of IbBcBa, IbB'cB'a, resp.

C1, C2 = the circumcenters of IcCaCb, IcC'aC'b, resp.

1. The circles (Q),(A1),(A2) concur at a point Da

The circles (Q),(B1),(B2) concur at a point Db

The circles (Q),(C1),(C2) concur at a point Dc

Properties of the points? (Perspectivity of DaDbDc with triangles ??)

2. The circumcenters A1,A2 - B1, B2 - C1, C2 are symmetric with center of symmetry the circumcenter Q of IaIbIc. (They lie on a conic with center Q)

3. The Triangles

3.1. ABC, Bounded by (AbAc, BcBa, CaCb)

3.2. ABC, Bounded by (A'bA'c, B'cB'a, C'aC'b)

3.3. Bounded by (AbAc, BcBa, CaCb), Bounded by (A'bA'c, B'cB'a, C'aC'b)

are perspective.

Perspectors?

Antreas P. Hatzipolakis, 24 April 2013