Altitudes and Orthic Triangles

Karyn Carson

Construct triangle ABC and its
circumcircle:

Next, construct the altitudes of
triangle ABC and mark their intersections on the circumcircle:

Then, construct the triangle that
is formed from those intersections:

If you construct the orthic
triangle (HIJ), you will see that it and the triangle formed by the altitudes
(EFG) of the original triangle (ABC) are similar. At least it looks that way. LetŐs investigate using information on parallel lines.

If I can prove that HI is parallel
to GF, and IJ is parallel to FE, and JH is parallel to EG, then the triangles
must be similar. To start
with, segment AB is a transversal of segments HI and GF. Then we need to look at the angles
formed by that transversal, BHI and GmH (alternate interior angles).

If a transversal crosses two lines
or segments that are parallel, then their alternate interior angles must be
equal. In this case they are, so
the segments must be parallel.

We can also measure the angles for
the other segments to prove that they are parallel:

and

What happens if itŐs a different
triangle. Will this proof still
work?

The alternate interior angles are still
congruent, so therefore the triangles are still similar.

What happens if we make our
original triangle obtuse? What
happens to our similar triangles?

When angle C becomes a right
angle, the orthocenter becomes the same point as C and we lose our orthic
triangle as well as the ÔaltitudeŐ triangle. If angle C becomes obtuse,

then the orthocenter leaves
triangle ABC, because two of the altitudes have also left the triangle (they
must be drawn outside the triangle to be perpendicular with the base).