Let $P$ be a point on the circumcircle of a triangle $A_{1}A_{2}A_{3}$, and let $H$ be the orthocenter of the triangle. The feet $B_{1},B_{2},B_{3}$ of the perpendiculars from $P$ to $A_{2}A_{3},A_{3}A_{1},A_{1}A_{2}$ lie on a line. Prove that this line bisects the segment $PH$.
Problem
Source: 4-th Taiwanese Mathematical Olympiad 1995
Tags: geometry, circumcircle, geometry unsolved
cvix
16.01.2007 20:20
You can find a proof in a book Coxeter, Greitzer "Geometry revisited" in the chapter "More on Simson lines" (or something like that).
N.T.TUAN
19.01.2007 15:57
cvix wrote: You can find a proof in a book Coxeter, Greitzer "Geometry revisited" in the chapter "More on Simson lines" (or something like that). Post a solution, please! Because I haven't got that book!
nayel
19.01.2007 16:41
its a very famous as well as useful book on geometry. you should collect it.
vittasko
21.01.2007 23:28
N.T.TUAN wrote: cvix wrote: You can find a proof in a book Coxeter, Greitzer "Geometry revisited" in the chapter "More on Simson lines" (or something like that). Post a solution, please! Because I haven't got that book! It is easy to prove that $QR\parallel P'H,$ where $QP' = QP$ and so, the line $QR$ $($ simson's line of $P$ $),$ bisects the segment $PH.$ kostas Vittas.
Attachments:
