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• Prove that the line joining the -circumcenter of a triangle to a vertex- is perpendicular to the line -joining the foot of altitudes of adjoining sides.- Some
Message 1 of 3 , Sep 30, 2007
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Prove that the line joining the -circumcenter of a triangle to a
vertex- is perpendicular to the line -joining the foot of altitudes of
Some sort of a HINT please....
• dear colleague There is a very simple response the orthic triangle is homothetic to the tangential one The line joining the -circumcenter of a triangle to a
Message 1 of 3 , Oct 1, 2007
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dear colleague
There is a very simple response
the orthic triangle is homothetic to the tangential one
The line joining the -circumcenter of a triangle to a
vertex- is perpendicular to the tangent line in that vetex to the
circumcircle, so to the parallel side of the orthic triangle

There are others proves using barycentrics coordinates

Best regards

Michel Garitte

--- In Hyacinthos@yahoogroups.com, "malisundaresan"
<malisundaresan@...> wrote:
>
> Prove that the line joining the -circumcenter of a triangle to a
> vertex- is perpendicular to the line -joining the foot of altitudes
of
> Some sort of a HINT please....
>
• ... altitudes of ... Let E, F be the feet of the altitudes through B, C upon AC, AB respectively, Let X be the midpoint of AH (H orthocenter) X is the center
Message 1 of 3 , Oct 1, 2007
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--- In Hyacinthos@yahoogroups.com, "malisundaresan"
<malisundaresan@...> wrote:
>
> Prove that the line joining the -circumcenter of a triangle to a
> vertex- is perpendicular to the line -joining the foot of
altitudes of
> Some sort of a HINT please....
>
Let E, F be the feet of the altitudes through
B, C upon AC, AB respectively,
Let X be the midpoint of AH (H orthocenter)
X is the center of the circle through A, E, H, F.
XE = XF
The nine-point circle (N) passes through E and F
NE = NF (each = R/2)
Follows NX is the perpendicular bisector of EF.
In the triangle AOH :
X, N being the midpoints of AH, OH we have
NX is parallel to OA.
Together with the fact that NX is perpendicular to EF,
it now follows that OA is perpendicular to EF