- 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 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

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:>

of

> Prove that the line joining the -circumcenter of a triangle to a

> vertex- is perpendicular to the line -joining the foot of altitudes

> adjoining sides.-

> Some sort of a HINT please....

> - --- In Hyacinthos@yahoogroups.com, "malisundaresan"

<malisundaresan@...> wrote:>

altitudes of

> Prove that the line joining the -circumcenter of a triangle to a

> vertex- is perpendicular to the line -joining the foot of

> adjoining sides.-

Let E, F be the feet of the altitudes through

> Some sort of a HINT please....

>

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

Vijayaprasad