Altitudes and the Orthocenter

By Sharon K. O’Kelley

 

The Problem

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 ….

                                                                                                                  Prove….

 

 

 

Proof 1

 

                                                                                                                    Prove ….

Since there are three different altitudes or height with three different corresponding bases, the area of triangle ABC can be expressed in three ways….

 

 

 

 

 

Note that within triangle ABC there are three smaller triangles – i.e., , , and  The areas of these triangles are….

 

 

 

 

The areas of the smaller triangles and of triangle ABC can now be expressed as ratios….

 

 

 

 

Consider that the three smaller triangles comprise triangle ABC; therefore, the sum of their areas will equal the area of triangle ABC. This fact allows for the following equation….

 

 

 

This equation can be manipulated to produce….

 

 

Using substitution, the desired result is obtained….

 

 

                    Q.E.D.

 

 

 

 

Proof 2

 

                                                                                                                    Prove….

 

Consider the result from Proof 1 that….

 

In triangle ABC, segments AH, BH, and CH are parts of their respective altitudes; therefore, they can be used to express the above equation using a variation of the segment addition postulate….

 

 

 

This equation can now be rewritten as….

 

 

 

 

This can now be manipulated to obtain the desired result….

 

            Q.E.D.

 

 

What if the Given Triangle is Obtuse? Will These Two Results Hold?

It was demonstrated in Proof 1 that when triangle ABC is acute that….

It follows then that….

This also makes sense because segment HE is a part of segment BE; therefore, it is less than the whole.

Now consider when triangle ABC is obtuse and the orthocenter is outside the triangle as shown in the diagram below.

Now segment BE is a part of segment HE therefore….

This would mean that….

 

Thus, the results from Proof 1 and thereby Proof 2 do not hold for the obtuse case.

 

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