14973Re: [EMHL] Four Circum Equilateral Triangles
- Mar 5, 2007Dear Quang!
The point P is isogonally conjugated to the infinite point of Euler line, i.e the common point of the circumcircle and the Neuberg cubic distinct from A, B, C and two circular points.
Dear All My Friends,
> Given triangle ABC with circumcircle (O). There is and exactly oneRecent Activity
> equilateral triangle inscribed in (O) taken A as one vertex. We denote its
> bottom line wrt A as La, intesection of La with sideline BC as A'. Similarly
> define Lb, B', Lc, C'. Three lines La, Lb, Lc bound one triangle A''B''C''.
> Please prove following results:
> 1. ABC and A''B''C'' are perspective with perspector P on circumcircle
> 2. A', B', C' are collinear on one line L cutting (O) at two point Q, R.
> 3. PQR are our fourth equilateral.
> Thank you and best regards,
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