Geometric nature of X
- Dear Mathlinkers,
1. ABC a triangle
2. I the incenter of ABC
3. A’B’C’ the I-cevian triangle of ABC
4. LMN the medial triangle of ABC
5. A’’ the point of intersection MI and B’C’, and circularly B’’, C’’
Prouve : A’’B’’C’’ and ABC are perspective at X
[Non-text portions of this message have been removed]
- X is the symmedian point (barycentric square of I).
If you change I by any other point P then you'll get that X is the the barycentric square of P.
--- In Hyacinthos@yahoogroups.com, Jean-Louis Ayme <jeanlouisayme@...> wrote:
> Dear Mathlinkers,
> 1. ABC a triangle
> 2. I the incenter of ABC
> 3. AâBâCâ the I-cevian triangle of ABC
> 4. LMN the medial triangle of ABC
> 5. Aââ the point of intersection MI and BâCâ, and circularly Bââ, Cââ
> Prouve : Â Â Â Â Â Â Â Â Â Â Â Â Â Â AââBââCââ and ABC are perspective at X
> [Non-text portions of this message have been removed]
- If you change LMN by cevian triangle of a point P, then X will be the isogonal conjugate of P wrt ABC.
Suppose, AA1 is the external angle-bisector of \angle BAC where A1 lies on BC. IM intersects AA1 at A2. Then note that, (l_a,IA1;B'C',BC)=-1. So (A2I;B"M)=-1. So (AA1,AI;AB",AM)=-1. So AB" is the isogonal conjugate of AM and similar for others. So done.