Orthocenter - The Point of concurrency of 3 Altitudes of a Triangle

Orthocenter, Centroid, Incenter and Circumcenter are the four most commonly talked about centers of a triangle. I have written a great deal about the Incenter, the Circumcenter and the Centroid in my previous posts. In this post, I will be specifically writing about the Orthocenter. 

Orthocenter is the point in the plane of a triangle where all three altitudes of the triangle intersect. An altitude of a triangle is the segment from a vertex perpendicular to the side opposite to the vertex, or perpendicular to the line that contains the opposite side. In the figure below, AD is the altitude from vertex A perpendicular to BC(or the line containing the side BC), the side sitting opposite of vertex A, of △ABC. 

Similarly, BE and DF are the other two altitudes of triangle ABC emanating from vertices B and C. All three altitudes intersect at point H - the Orthocenter of the triangle. 

In any triangle, the three altitudes are always concurrent(intersecting at a single point), and so the Orthocenter exists in the plane of every triangle. For an acute angled triangle, the Orthocenter will be sitting inside of the triangle, like in the case of △ABC above. 

In an obtuse angled triangle, the Orthocenter will be present outside of the triangle. And for a right angled triangle, the location of the Orthocenter is exactly at the vertex where 90° angle is formed. 

Using converse of ceva’s theorem it can be proved that the three altitudes are concurrent in acute and obtuse triangles. But it is kind of obvious to see why the three altitudes of a right angled triangle will have to intersect at a single point, and why that point happens to be the vertex of the right angle. 

This is simply because the two sides in a right triangle are perpendicular to each other. So the altitudes to those two sides overlap on the sides as seen in the figure above. The altitude of BC overlaps the side AB and the altitude of side AB overlaps the side BC. These two altitudes meet at the vertex C where there is 90° angle. And the third altitude to the hypotenuse starts from the vertex C. So C is clearly the point where all three altitudes meet. 

We can also prove this by converse of ceva’s theorem, something that I have already done in my previous post. 

Let’s look at the proof for an acute angled triangle using the converse of Ceva’s Theorem. 

Let △ABC be an acute angled triangle. Points D, E and F are where the altitudes from the vertices A, B and C meet the sides. According to the converse of Ceva’s theorem, in order for the three altitudes to be concurrent the following must be true : 

\(\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}\) = 1

The above relation is what we need to prove. 

For clarity purposes, we are looking at the three altitudes in three separate figures which are just three copies of the same triangle. Each altitude divides the original triangle ABC into two smaller right angled triangles. 

In △ABD, 

cos B = \(\frac{BD}{AB}\)  

BD = AB cos B

Similarly, in △ADC, 

DC = AC cos C

In △BEA,

EA = AB cos A

In △BEC,

CE = BC cos C

In △CFA, 

AF = AC cos A

In △CFB, 

FB = BC cos B

So, L.H.S = \(\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}\)

= \(\frac{AC \cdot \cos A}{BC \cdot \cos B} \cdot \frac{AB \cdot \cos B}{AC \cdot \cos C} \cdot \frac{BC \cdot \cos C}{AB \cdot \cos A}\)

= 1 = R.H.S

Hence by converse of ceva’s theorem the three altitudes in an acute angled triangle are concurrent. 

The proof for an obtuse angled triangle works on the same lines. 

(△ABC is shaded in colour) 

To prove that : \(\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}\) = 1

In △AEB,  

EA = AB cos(π - A)

Similarly, in triangles BEC, ADB, ADC, AFC, BFC

CE = BC cos C

BD = AB cos B

DC = AC cos C

AF = AC cos(π - A)

FB = BC cos B

So, L.H.S = \(\frac{AC \cdot \cos(\pi - A)}{BC \cdot \cos B} \cdot \frac{AB \cdot \cos B}{AC \cdot \cos C} \cdot \frac{BC \cdot \cos C}{AB \cdot \cos(\pi - A)}\)

                = 1 = R.H.S

Circumcenter is the center of the circle which circumscribes the triangle. Incenter is the center of the Incircle. Incircle is the circle of greatest possible radius inside the triangle. What about Orthocenter? Is it also the center of some circle? 

Well, in a way yes, but the circle doesn’t directly involve the principal triangle. When we join the foot of the three altitudes, we get another triangle inside of the principal or original triangle. This inside triangle is called the Orthic triangle. 

The Orthocenter is the incenter of the orthic triangle. The altitudes of the original triangle are the three angle bisectors of the orthic triangle. 

The Orthocenter is also the center of the circumcircle of the anticomplementary triangle of the original triangle. That’s quite a lot. 

Let △PQR be an anti-complementary triangle of the main triangle △ABC. The three sides of the anti-complementary triangle △PQR pass through the vertices A, B and C of the main triangle, and are parallel to the sides opposite to the vertices of the main triangle.

The Orthocenter of the main triangle is the center of the circumcircle of the anti-complementary triangle of the main triangle.