Advanced Triangle Centers: Orthocenter

The orthocenter of $\Delta ABC$ is the point where the altitudes of $\Delta ABC$ concur.

PROOF OF CONCURRENCE:

Suppose that $\Delta ABC$ is the Medial Triangle of $\Delta HGI.$ Since $AD$, $BE$, and $CF$ are perpendicular bisectors of the sides of $\Delta HGI$, they concur at the circumcenter. Hence, the altitudes of $\Delta ABC$ concur at the orthocenter.

$\blacksquare$

CONCLUSION: The orthocenter as well as many other triangle centers show up often in mathematics. It is quite useful to know how to work with them.

Drawing a Diagram : Constructing the Orthocenter

This is fairly easy.

Step 1 : Draw a nice, big triangle.

Step 2 : Draw the A-altitude and B-altitude. (I'm sure you can approximate very close to a right angle, with the help of a ruler.)

Step 3 : The intersection is the orthocenter of the triangle.

and we're done.

The three altitudes of a triangle meet at a single point, known as the orthocenter. This is a fact that is far from obvious. One problem with proving this statement is that the altitudes can meet either inside the triangle, outside the triangle, or on the triangle itself. In addition, sometimes, the altitudes as we normally draw them will intersect as desired, whereas sometimes we need to extend the altitudes to make them meet (so perhaps we should qualify the above statement as follows: the lines containing the three altitudes of a triangle meet at the orthocenter).

A common method of proving that three or more lines are concurrent is to define the intersection of any two of them, and then to prove that the other lines pass through the point we have defined. For this proof, let the triangle be $\triangle ABC$, and let the altitude from $A$ to $BC$ and the altitude from $B$ to $AC$ be $H$ (the orthocenter is typically denoted $H$ because it is the intersection of the heights of the triangle). We must prove that the altitude from $C$ to $AB$ passes through $H$.

A very elegant and clean way of proving this is to use vectors. The motivation for using vectors is that vectors have a very nice condition for perpendicularity; in addition, we no longer have to prove the theorem for multiple configurations (e.g. the intersection point lies inside or outside the triangle). The use of vectors allows full generality.

Since $H$ lies on the altitude from $A$ to $BC$, we have: $$\vec{AH}\perp\vec{BC}$$ Two vectors are perpendicular if and only if their dot product is zero: $$\vec{AH}\cdot\vec{BC} = 0$$ Now, we do some algebra: $$(\vec{H}-\vec{A})\cdot(\vec{C}-\vec{B}) = 0$$ $$\vec{H}\cdot\vec{C}-\vec{H}\cdot\vec{B}-\vec{A}\cdot\vec{C}+\vec{A}\cdot\vec{B} = 0$$

Since $H$ lies on the altitude from $B$ to $AC$, we have: $$\vec{BH}\perp\vec{AC}$$ Performing the same process as above, we reach the following: $$\vec{H}\cdot\vec{C}-\vec{H}\cdot\vec{A}-\vec{B}\cdot\vec{C}+\vec{A}\cdot\vec{B} = 0$$

Subtracting the first equation from the second equation, we get: $$\vec{H}\cdot\vec{B}+\vec{A}\cdot\vec{C}-\vec{H}\cdot\vec{A}-\vec{B}\cdot\vec{C} = 0$$ Factoring the left side, $$(\vec{H}-\vec{C})(\vec{B}-\vec{A}) = 0$$ $$\vec{CH}\cdot\vec{AB}=0$$ Therefore, $\vec{CH}\perp\vec{AB}$, so $H$ lies on the line containing the altitude from $C$ to $AB.$ It follows that the three altitudes of a triangle must concur, and the proof is complete. We call the point $H$ the orthocenter of $\triangle ABC.$