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This essential trains for: AMC-10, AMC-12, AIME.

An altitude (height) of a triangle is a segment that is perpendicular to a side and passes through the vertex that is opposite to that side.

To prove that the three altitudes of a triangle are concurrent, we will show that the altitudes of a triangle are collinear with the perpendicular bisectors of another triangle. Since the perpendicular bisectors of the sides of a triangle are concurrent (see proof in other lesson), the altitudes of the related triangle are also concurrent.

In triangle ABC draw parallels to each side through the opposite vertex: A larger triangle is formed by the intersection of these lines. Denote it by A'B'C'.

By construction:  Hence, ABA'C is a parallelogram and B'CBA is also a parallelogram.

Therefore,:  By transitivity, and hence C is a midpoint of A'B'.

The altitude CP is perpendicular onto AB but, due to parallelism, it is also perpendicular onto A'B' at the midpoint C. Therefore, an altitude of the triangle ABC is also a perpendicular bisector in the triangle A'B'C'. The same argument can be proven similarly for each altitude.

Therefore, because the perpendicular bisectors of the triangle A'B'C' are concurrent, so are the altitudes of the triangle ABC.

The point of intersection of the altitudes of a triangle is called the orthocenter of the triangle.

Note: The orthocenter of a triangle may lie outside the triangle (if the triangle is obtuse). The proof guarantees that the altitudes have a common intersection. However, if one of the angles of the triangle is obtuse, two of the altitudes do not intersect the interior of the triangle - hence, the orthocenter must be a point that is outside the triangle.

Note: The orthocenter of a right triangle is the vertex of the right angle.