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Re: feet of altitudes and angles
Posted:
Nov 27, 1997 8:29 PM
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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 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:
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'=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'.
Regards,
Floor van Lamoen
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