Baltic 2009 . It`s just Ceva (in sinus form) .

any solution

Dear Mathlinkers, this is a variation of de Villiers M., A dual to Kosnita's theorem, Mathematics & Informatics Quarterly 6 (3) (sept. 1996) 169-171 Sincerely Jean-Louis

$\textbf{Proof :}$ Consider external homothetic center $H_{12}$ of in-circles $(I_1), (I_2)$. It's well known that $H\in AB$. So $I_1I_2\cap AB = H_{12}$. Similarly define points $H_{23}, H_{31}$. By $\textbf{Monge's theorem}$ we get that $H_{12}, H_{23}, H_{31}$ lies on same line. So by $\textbf{Desargues theorem}$ we get that triangles $ABC$ and $I_1I_2I_3$ are perspective. $\Box$

euclideangeometry wrote: The point $H$ is the orthocentre of a triangle $ABC$, and the segments $AD,BE,CF$ are its altitudes. The points $I_1,I_2,I_3$ are the incentres of the triangles $EHF,FHD,DHE$ respectively. Prove that the lines $AI_1,BI_2,CI_3$ intersect at a single point. $\angle HEF=90^{\circ}-B$ and $\angle HFE=90^{\circ}-B$.So $\frac {I_1F}{I_1E}=\frac{\sin(45^{\circ}-B/2)}{\sin(45^{\circ}-C/2)}$ by applying dine rule in triangle $I_1EF$. Now note that $\frac{\sin FAI_1}{\sin I_1AE}=\frac{\sin(\angle I_1FA)}{\sin \angle I_1EA}\frac{I_1F}{I_1E}=\frac{\sin 45^{\circ}+C/2}{\sin (45^{\circ}+B/2 )}\frac{\sin (45^{\circ}-B/2)}{\ sin (45^{\circ}-C/2 )}$. So by symmetry and trig ceva the conclusion follows..