Altitudes and Orthocenters
by
Ana Kuzle
I began by constructing an arbitrary triangle ABC. Then I constructed the orthocenter of triangle ABC. That is, the common intersection of the three lines containing the altitudes, and labeled H. Afterwards, I constructed, using orthocenter script tool, orthocenters of triangles BCH, CAH and ABH. From the picture bellow it can be seen that the orthocenters of the triangles BCH, CAH and ABH are A, B and C, respectively. Click here for GSP file.
The following picture shows circumcircles of triangles ABC, CAH, BCH and ABH. In it we can see that S, J, K and L are circumcenters of those triangles respectively. Click here to see the GSP file.
What can we see from the picture? Well, first of all it can be seen that the orthocenter H of triangle ABC is the intersection of the circumcircles of the triangles CAH, BCH and ABH.
Take a look of this! For instance, if B is moved to the orthocenter H, the corresponding triangle would be CAH. Why is this so? Well, this overlapping occurs because triangle ABC overlaps the corresponding triangle and therefore produces the same circumcenter and circumcircle.
I continued my investigation by construct any acute triangle ABC and its circumcircle. Then, constructed the three altitudes AD, BE, and CF; extended each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.
Moving each of the vertices did not change the sum of those rations obtaining acute triangle all time, naturally. Click here for GSP file.
Proof.
where
(ASA triangle congruence)
Thus, in order to prove original problem we have to prove that
We know that
and by looking at the following ratios of the triangle areas
we get
In the end given an acute triangle ABC, I constructed the orthocenter H. Let points D, E, and F be the feet of the perpendiculars from A, B, and C respectively. Measuring shows that
Moving the vertices does not change that sum of the ratios as long as triangle is acute. Click here for GSP file.
Proof. Similarly as above we get that
