directed segment,angle

Leonhard Euler

When we have to show $D,E,F$ are collinear, where $D,E,F$ are points on side of $BC,CA,AB$ respectively, then it is suffice to show $\frac {\overline{DB}}{\overline{CD}}\frac {\overline{EA}}{\overline{BE}}\frac {\overline{CF}}{\overline{AF}} = - 1$ . Sometimes it is easy to show $\frac {DB}{CD}\frac {EA}{BE}\frac {CF}{AF} = 1$ but not easy to show $\frac {\overline{DB}}{\overline{CD}}\frac {\overline{EA}}{\overline{BE}}\frac {\overline{CF}}{\overline{AF}} = - 1$ . I want to know skill of proving this.

Let $ABC$ be a triangle and $DEF$ be cevian triangle. Let $X$ be the inersection of $BC$ and tangent line at $A$ on circumcircle of triangle $AEF$ and define $Y,Z$ similarly. Prove that $D,E,F$ are collinear.

In this problem, It is not hard to show $\frac {XB}{CX}\frac {YA}{BY}\frac {CZ}{AZ} = 1$ since $\frac {XB}{CX} = \frac {AB\cdot \sin BAX}{AC\cdot \sin CAX} = \frac {AB}{AC}\frac {\sin AEF}{\sin AFE} = \frac {AB}{AC}\frac ...$ and similarly hold for $\frac {YA}{BY},\frac {CZ}{AZ}$ . But how can we show $\frac {\overline{XB}}{\overline{CX}}\frac {\overline{YA}}{\overline{BY}}\frac {\overline{CZ}}{\overline{AZ}} = - 1$ ? I think it need conception of directed angle.

**Posted:** Tue Feb 17, 2009 3:55 am