Let E,F,G be the excenters of triangle ABC. How can I prove that the orthic triangle (triangle with vertices at the feet of the altitudes of the bigger triangle) of EFG is ABC?
Thanks for any help.
Let E be the center af the excircle wich is tangent to BC. Then AE is the bisector of angle A. AF and AG are bisectors of the two exterior, opposite, angles of A, so F, A, G are colinear.
Now, $\displaystyle \widehat{BAE}=\widehat{EAC}$ and $\displaystyle \widehat{FAC}=\widehat{GAB}$. Then $\displaystyle EA\perp FG$.
In the same way, $\displaystyle FB\perp GE, \ GC\perp EF$
So ABC is the orthic triangle.