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 Math Help > Geometry > Polygons and Triangles > Triangle Centers > Orthocenter

The Orthocenter of a triangle is the point formed by the intersection of its three altitudes (or their extensions).

 Orthocenter, concurrency of the three altitudes

### Concurrence of the altitudes

The three altitudes have to meet in a single point because...

Cut-the-knot lists a dozen or more proofs, but the simplest to understand (for me) uses Carnot's Theorem this way:

Letting AA', BB', and CC' be the altitudes of ABC, we see that each altitude is a leg of two right triangles, so that

CC'2 + AC'2 = AC2,
-CC'2 - BC'2 = -BC2,
AA'2 + BA'2 = AB2,
-AA'2 - CA'2 = -AC2,
BB'2 + CB'2 = BC2,
-BB'2 - AB'2 = -AB2

Now, just as we did to prove Carnot's theorem, we simply add up the six equations to get

AC'2 - BC'2 + BA'2 - CA'2 + CB'2 - AB'2 = 0, and then by Carnot's theorem, the three altitudes concur on a point.

### Barycentric coordinates of the orthocenter

( (a^2+b^2-c^2)(c^2+a^2-b^2), (b^2+c^2-a^2)(a^2+b^2-c^2), (c^2+a^2-b^2)(b^2+c^2-a^2) )

You can think of the barycentric coordinates of the "weights" of the vertices, so the weighted average of the vertices is the circumcenter.  You can calculate the circumcenter from the barycentric coordinates by multiplying each vertex (vector) by the corresponding barycentric coordinate, adding the results, and then dividing by the sum of the barycentric coordinates.

If we represent A, B, and C on the coordinate plane by (a,b), (c,d), (e,f) and replace the squares of the lengths of the sides appropriately, we can calculate the orthocenter (h,k) as

h = ( a*(((c-e)^2+(d-f)^2)+((a-e)^2+(b-f)^2)-((a-c)^2+(b-d)^2))* (((a-c)^2+(b-d)^2)+((c-e)^2+(d-f)^2)-((a-e)^2+(b-f)^2))+ c*(((a-e)^2+(b-f)^2)+((a-c)^2+(b-d)^2)-((c-e)^2+(d-f)^2))* (((c-e)^2+(d-f)^2)+((a-e)^2+(b-f)^2)-((a-c)^2+(b-d)^2))+ e*(((a-c)^2+(b-d)^2)+((c-e)^2+(d-f)^2)-((a-e)^2+(b-f)^2))* (((a-e)^2+(b-f)^2)+((a-c)^2+(b-d)^2)-((c-e)^2+(d-f)^2)) ) / ( (((c-e)^2+(d-f)^2)+((a-e)^2+(b-f)^2)-((a-c)^2+(b-d)^2))* (((a-c)^2+(b-d)^2)+((c-e)^2+(d-f)^2)-((a-e)^2+(b-f)^2))+ (((a-e)^2+(b-f)^2)+((a-c)^2+(b-d)^2)-((c-e)^2+(d-f)^2))* (((c-e)^2+(d-f)^2)+((a-e)^2+(b-f)^2)-((a-c)^2+(b-d)^2))+ (((a-c)^2+(b-d)^2)+((c-e)^2+(d-f)^2)-((a-e)^2+(b-f)^2))* (((a-e)^2+(b-f)^2)+((a-c)^2+(b-d)^2)-((c-e)^2+(d-f)^2)) )

which simplifies to

h = ( (d-f)b^2+(f-b)d^2+(b-d)f^2+ab(c-e)+cd(e-a)+ef(a-c) ) / (bc+de+fa-cf-be-ad), and
k = ( (e-c)a^2+(a-e)c^2+(c-a)e^2+ab(f-d)+cd(b-f)+ef(d-b) ) / (bc+de+fa-cf-be-ad)

### Other factoids about the orthocenter

The orthocenter and circumcenter are isogonal conjugates of one another.

If H is the orthocenter of triangle ABC, then...

A is the orthocenter of triangle HBC,
B is the orthocenter of triangle HCA, and
C is the orthocenter of triangle HAB.

Together, A, B, C, and H are said to represent an Orthocentric System  Why is this true?  Consider the six lines formed by points A, B, C, and H.  They form three pairs of perpendicular lines:

AB is perpendicular to CH,
AC is perpendicular to BH, and
BC is perpendicular to AH

Moreover, if you pick any two of the four points A, B, C, and H, these two points determine a line, and the two points not picked determine its perpendicular.  Now, let P, Q, and R be any three points selected from among A, B, C, and H to make triangle PQR, and let S be the fourth point, also selected from among A, B, C, and H.  The PQ and RS are perpendicular, so RS is one of the three altitudes of PQR.  Similarly, PS and QS are altitudes of PQR, so all three altitudes pass through S, making S the orthocenter of PQR.

Let ABCH be an orthocentric system, and without loss of generality, assume H is in the interior of triangle ABC.  Let HA, HB, and HC represent the three intersections of perpendicular lines in the system.  Now, for your amusement, observe the following cyclic quadrilaterals:

 cyclic quadralateral diameter of its circumcircle HHCBHA HB HHBAHC HA HHACHB HC ABHAHB AB BCHBHC BC CAHCHA CA

Interestingly, of the seven points A, B, C, H, HA, HB, and HC, for each of the six sets of collinear points, the remaining four are the vertices of a cyclic quadrilateral.

The orthic triangle is the triangle formed by the feet of the altitudes, HA, HB, and HC.  The incenter of the orthic triangle HAHBHC is the orthocenter of ABC.

### Internet references

Cut-the-knot: Triangle Altitudes and Orthocenter has a dozen proofs that the altitudes are concurrent, establishing many fun facts along the way!

Mathworld: Orthocentric System -- more fun facts about orthocenters and their relationship to the 9-point circle.

### Related pages in this website

Other triangle centers: Circumcenter, Incenter, Orthocenter, Centroid.  The Orthocenter and Circumcenter of a triangle are isogonal conjugates, and the Incenter is its own isogonal conjugate.

Summary of geometrical theorems summarizes the proofs of concurrency of the lines that determine these centers, as well as many other proofs in geometry.

Barycentric Coordinates, which provide a way of calculating these triangle centers (see each of the triangle center pages for the barycentric coordinates of that center).

Carnot's Theorem -- AC'2 - BC'2 + BA'2 - CA'2 + CB'2 - AB'2 = 0 iff perpendiculars from A', B', and C' concur on a point.

Triangle Centers -- a summary of the different "centers" of triangles.

The Orthocenter and Circumcenter of a triangle are isogonal conjugates.

The webmaster and author of this Math Help site is Graeme McRae.