Re: Triangle altitudes intersect-

The altitudes of a triangle ABC are the bisectors of its altitude triangle, the triangle with the feet of the altitudes as its vertices. So they are concurrent; their intersection point G is named the orthocentre of ABC.

The following proof was quoted from "P. Wijdenes: Vlakke Meetkunde voor Voortgezette Studie. Wolters-Noordhoff 1968".

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Let ABC be an acute triangle and let D, E, F be the feet of the altitudes from A, B, C.
Then the rectangular triangles ACD and BCE are similar, and there exists a ratio k such that CD = k.CA and CE = k.CB.
Therefore the triangles ABC and DEC are similar. Likewise, ABC is similar to DBF and to AEF.

So AB and DE run antiparallel with respect to AC and BC. Likewise, BC and EF wrt AB and AC, and AC and DF wrt AB and BC, and there are at points D, E, F pairs of equal angles between the sides of the original triangle ABC and its altitude triangle DEF. This proves that the sides AB, BC, CA of ABC are the outer bisectors of DEF.
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Obviously the altitudes are the inner bisectors. Strangely enough this theorem is not mentioned as such in Wijdenes's book.

Minor alterations in the above apply for obtuse triangles ABC.

The way back is as follows:
Draw all the inner and outer bisectors of the original triangle DEF.
The inner bisectors intersect at a point G; it happens three times that two of the outer bisectors and one of the inner bisectors intersect in point A, point B, point C, respectively.

The original triangle DEF is the altitude triangle of ABC. The point G is the orthocentre of ABC.

The circumcircle of DEF is the nine-point circle of ABC.

See http://en.wikipedia.org/wiki/Triangle and http://en.wikipedia.org/wiki/Nine-point_circle

Ciao: Johan E. Mebius


gmarkowsky@xxxxxxxxx wrote:

Hello all,

I was thinking about the fact that the three altitudes of a triangle
meet at a point. The proof I thought of was just to put the three
points of the triangle at (-a,0), (b,0), and (0,h)(for the picture you
can imagine a,b > 0), then just calculate the equations of the
altitudes from the first two points and see where they intersect x=0.
This works, of course, but there must be a more elegant way to do. Like
how would the ancient Greeks have done it? They were definitely aware
of this theorem. Anyone how to prove it in a better way?

Greg



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