Question

let triangle ABC be a triangle in which all three interior angles are acute and let A'B'C' be the orthic triangle.

a.) Prove that the altitudes of triangle ABC are the angle bisectors of triangle A'B'C'.

b.) Prove the orthocenter of triangle ABC is the incenter of traingle A'B'C'.

c.) Prove that A is the A' -excenter of triangle A'B'C'.

Answer 1

Altitudes and the Orthic Triangle of Triangle ABC:

Given a triangle ABC with acute angles, let A*, B*, C* be the feet of the altitudes of the triangle: A*, B*, C* are points on the sides of the triangle so that AA* BB*, CC* are altitudes.

Then we have proved earlier that the altitudes are concurrent at a point H.  (The proof used the relationship between the perpendicular bisectors of the sides of a triangle and the altitudes of its midpoint triangle).

The orthic triangle of ABC is defined to be A*B*C*.  This triangle has some remarkable properties that we shall prove:

1. The altitudes and sides of ABC are interior and exterior angle bisectors of orthic triangle A*B*C*, so H is the incenter of A*B*C* and A, B, C are the 3 ecenters (centers of escribed circles).
2. The sides of the orthic triangle form an "optical" or "billiard" path reflecting off the sides of ABC.
3. From this it can be proved that the orthic triangle A*B*C* has the smallest perimeter of any triangle with vertices on the sides of ABC.

b)Orthocenter of ABC is the incenter of A'B'C': The orthocenter H of ABC is the incenter of A*B*C*, and A, B and C are the ecenters of A*B*C*.  Thus four circles tangent to lines A*B*, B*C*, C*A* can be constructed with centers A, B, C, H.

Relation between the Orthocenter and the Circumcircle

The triangle ABC can be inscribed in a circle called the circumcircle of ABC.  There are some remarkable relationships between the orthocenter H and the circumcircle.

The altitude line CC* intersects the circumcircle in two points.  One is C.  Denote the other one by C**