Annie Fetter wrote: > > My grandmother, who used to be a math teacher, read this problem long ago > and has always wondered about a proof. (She was never 100% sure that it > was really true until I modelled the problem in Sketchpad for her. Now she > is a believer.) I haven't tried to prove it yet, but told her I would > throw it out to the wolves and see what people came up with. > > Prove that the angles formed by connecting the feet of the altitudes of a > triangle are themselves bisected by the altitudes.
Your grandmother was right when you take an acute triangle. When you have an obtuse triangle you should change the angle-bisectors of two angles in the outer angle-bisectors. Two sides do function as the 'normal' bisectors then. Of course in a right triangle the triangle of the feet is degenerate.
Back to the acute case, let A'B'C' be the feet of A, B and C:
A C' B
Since <BB'C = <BC'C = 90 there is a circle passing through BC'B'C'. In this circle we can see that <CC'B'=<CBB'=90-C. In a similar way we can see that <CC'A'=90-C. Hence CC' bisects angle A'C'B'. Of course the same argumentation holds for BB' and AA'.