# Orthocenter

The **orthocenter** of a triangle is the intersection of the triangle's three altitudes. It has several important properties and relations with other parts of the triangle, including its circumcenter, incenter, area, and more.

The orthocenter is typically represented by the letter $H$.

## Finding the Orthocenter

The location of the orthocenter depends on the type of triangle. If the triangle is acute, the orthocenter will lie within it. If the triangle is obtuse, the orthocenter will lie outside of it. Finally, if the triangle is right, the orthocenter will be the vertex at the right angle.

Because the three altitudes always intersect at a single point (proof in a later section), the orthocenter can be found by determining the intersection of any two of them. This is especially useful when using coordinate geometry since the calculations are dramatically simplified by the need to find only two equations of lines (and their intersection).

## A triangle has vertices at $A=(0,0), B=(14,0)$, and $C=(5,12)$. What are the coordinates of the orthocenter?

The easiest altitude to find is the one from $C$ to $AB$, since that is simply the line $x=5$. The next easiest to find is the one from $B$ to $AC$, since $AC$ can be calculated as $y=\frac{12}{5}x$. A line perpendicular to $AC$ is of the form $y=-\frac{5}{12}x+b$, for some $b$, and as this line goes through $(14,0)$, the equation of the altitude is $y=-\frac{5}{12}x+\frac{35}{6}$.

Finally, the intersection of this line and the line $x=5$ is $\left(5,\frac{15}{4}\right)$, which is thus the location of the orthocenter. $_\square$

## Proof of Existence

The most natural proof is a consequence of Ceva's theorem, which states that $AD, BE, CF$ concur if and only if $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=1,$ where $D, E, F$ are the feet of the altitudes.

Note that $\triangle BFC \sim \triangle BDA$ and, similarly, $\triangle AEB \sim \triangle AFC, \triangle CDA \sim \triangle CEB$. Therefore,

$\frac{BF}{BD} = \frac{BC}{BA}, \frac{AE}{AF} = \frac{AB}{AC}, \frac{CD}{CE} = \frac{CA}{BC}.$

Multiplying these three equations gives us

$\frac{BF}{BD} \cdot \frac{AE}{AF} \cdot \frac{CD}{CE} = \frac{BC}{BA} \cdot \frac{AB}{AC} \cdot \frac{CA}{BC} = 1.$

Rearranging the left-hand side gives us

$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.$

Therefore, the three altitudes coincide at a single point, the orthocenter.

## Properties

*For convenience when discussing general properties, it is conventionally assumed that the original triangle in question is acute. The same properties usually apply to the obtuse case as well, but may require slight reformulation.*

Interestingly, the three vertices and the orthocenter form an **orthocentric system**: any of the four points is the orthocenter of the triangle formed by the other three.

An incredibly useful property is that the reflection of the orthocenter over any of the three sides lies on the circumcircle of the triangle. There is a more visual way of interpreting this result: beginning with a circular piece of paper, draw a triangle inscribed in the paper, and fold the paper inwards along the three edges. The three arcs meet at the orthocenter of the triangle.[1]

This result has a number of important corollaries. The most immediate is that the angle formed at the orthocenter is supplementary to the angle at the vertex:

$\angle ABC+\angle AHC = \angle BCA+\angle BHA = \angle CAB+\angle CHB = 180^{\circ}$

Another follows from power of a point: the product of the two lengths the orthocenter divides an altitude into is constant. More specifically,

$AH \cdot HD = BH \cdot HE = CH \cdot HF$

Similarly,

$\begin{aligned} AD \cdot DH &= BD \cdot CD\\ BE \cdot EH &= AE \cdot CE\\ CF \cdot FH &= AF \cdot BF. \end{aligned}$

The application of this to a right triangle warrants its own note:

If the altitude from the vertex at the right angle to the hypotenuse splits the hypotenuse into two lengths of $p$ and $q$, then the length of that altitude is $\sqrt{pq}$.

Another corollary is that the circumcircle of the triangle formed by any two points of a triangle and its orthocenter is congruent to the circumcircle of the original triangle. This is because the circumcircle of $BHC$ can be viewed as the Locus of $H$ as $A$ moves around the original circumcircle.

Finally, this process (remarkably) can be reversed: if *any* point on the circumcircle is reflected over the three sides, the resulting three points are collinear, and the orthocenter always lies on the line connecting them.

On a somewhat different note, the orthocenter of a triangle is related to the circumcircle of the triangle in a deep way: the two points are isogonal conjugates, meaning that the reflections of the altitudes over the angle bisectors of a triangle intersect at the circumcenter of the triangle.

Another important property is that the reflection of orthocenter over the midpoint of any side of a triangle lies on the circumcircle and is diametrically opposite to the vertex opposite to the corresponding side.

## Orthic Triangle

The triangle formed by the feet of the three altitudes is called the **orthic triangle**. It has several remarkable properties. For example, the orthocenter of a triangle is also the incenter of its orthic triangle. Equivalently, the altitudes of the original triangle are the angle bisectors of the orthic triangle.

The sides of the orthic triangle have length $a\cos A, b\cos B$, and $c\cos C$, making the perimeter of the orthic triangle $a\cos A+b\cos B+c\cos C$. The orthic triangle has the smallest perimeter among all triangles that could be inscribed in triangle $ABC$.

Kelvin the Frog lives in a triangle $ABC$ with side lengths 4, 5 and 6. One day he starts at some point on side $AB$ of the triangle, hops in a straight line to some point on side $BC$ of the triangle, hops in a straight line to some point on side $CA$ of the triangle, and finally hops back to his original position on side $AB$ of the triangle. The smallest distance Kelvin could have hopped is $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. What is $m+n$?

The circumcircle of the orthic triangle contains the midpoints of the sides of the original triangle, as well as the points halfway from the vertices to the orthocenter. This circle is better known as the nine point circle of a triangle.

The orthic triangle is also homothetic to two important triangles: the triangle formed by the tangents to the circumcircle of the original triangle at the vertices (the **tangential triangle**), and the triangle formed by extending the altitudes to hit the circumcircle of the original triangle.

## See Also

## References

[1] *Orthocenter curiousities*. Retrieved January 23rd from http://untilnextstop.blogspot.com/2010/10/orthocenter-curiosities.html

**Cite as:**Orthocenter.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/triangles-orthocenter/