Investigation of Orthocenter and Altitudes

by

Nathan Wisdom

The ORTHOCENTER (H) of a triangle is the common intersection of the three lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line containing the opposite side of the triangle.

We begin this investigation with the following set of constructions

1. Construct any
triangle ABC.

2. Construct the Orthocenter H of triangle ABC.

3. Construct the Orthocenter of triangle HBC.

4. Construct the Orthocenter of triangle HAB.

5. Construct the Orthocenter of triangle HAC.

6. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.

**Conjectures and Observations**

- The Orthocenters of triangle HBC, HAC, and HAB are the vertices of triangle ABC respectively.

- The Circumcircles of triangles ABC, HBC, HAB, and HAC are congruent.

- The triangle formed by the centers of the circumcircles of triangles HBC, HAB, and HAC is congruent to triangle ABC.

- If any vertex of the triangle ABC is move to where the orthocenter H is located, you will get a right angle triangle, where H is the right angle. Thus the new orthocenter of the triangle is a vertex of the triangle.

- The Orthocenter of triangle ABC is the circumcenter of the triangle formed by the centers of the circumcircles of triangle HBC, HAB and HAC.

NOW SOLVE This... Prove that triangle ABC is congruent to triangle EFG

The internal angle bisectors of triangle ABC are extended to meet the circumcircle at points L, M, and N, respectively. Find the angles of triangle LMN in terms of the angles A, B, and C.

Proof

Does your result hold only for acute triangles? From The GSP construction it is easy to see that the result holds not only for acute triangles.

Given triangle ABC. Construct the Orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectfully. Prove:

1=(HF/CF)+ (HD/AD)+ (HE/BE)

Now Solve This... Prove that

2=(AH/AD)+ (BH/BE)+ (CH/CF)