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Re: feet of altitudes and angles
Posted:
Nov 27, 1997 8:29 PM


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.
Hi Annie,
Your grandmother was right when you take an acute triangle. When you have an obtuse triangle you should change the anglebisectors of two angles in the outer anglebisectors. 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:
C
B' A'
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'=90C. In a similar way we can see that <CC'A'=90C. Hence CC' bisects angle A'C'B'. Of course the same argumentation holds for BB' and AA'.
Regards, Floor van Lamoen



