The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted $H$. [asy] size(15cm); markscalefactor = 0.01; dot((0,0)); dot((4,0)); dot((3,3)); dot((3,0)); dot((2,2)); dot((3,1)); dot((3.6,1.2)); draw((0,0)--(4,0)--(3,3)--cycle); draw((0,0)--(3.6,1.2),red); draw((2,2)--(4,0),red); draw((3,0)--(3,3), red); draw(rightanglemark((0,0),(3,0),(3,1))); draw(rightanglemark((4,0),(3.6,1.2),(0,0))); draw(rightanglemark((0,0),(2,2),(4,0))); label("Orthocenter",(2.85,1.1),W); [/asy] The lines highlighted are the altitudes of the triangle, they meet at the orthocenter.

Proof of Existence

Note: The orthocenter's existence is a trivial consequence of the trigonometric version of Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful.

[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), A1 = (B+C)/2, O = circumcenter(A,B,C), G = (A+B+C)/3, H = 3*G-2*O; draw(A--B--C--cycle); draw(A--G--H--cycle); draw(A1--G--O--cycle); label("A",A,(0,1));label("B",B,(0,-1));label("C",C,(0,-1));label("G",G,(1,-1));label("H",H,(0,-1));label("O",O,(-1,1));label("$A'$",A1,(0,-1));dot(H); [/asy] Consider a triangle $ABC$ with circumcenter $O$ and centroid $G$. Let $A'$ be the midpoint of $BC$. Let $H$ be the point such that $G$ is between $H$ and $O$ and $HG = 2 GO$. Then the triangles $AGH$, $A'GO$ are similar by side-angle-side similarity. It follows that $AH$ is parallel to $OA'$ and is therefore perpendicular to $BC$; i.e., it is the altitude from $A$. Similarly, $BH$, $CH$, are the altitudes from $B$, ${C}$. Hence all the altitudes pass through $H$. Q.E.D.

This proof also gives us the result that the orthocenter, centroid, and circumcenter are collinear, in that order, and in the proportions described above. The line containing these three points is known as the Euler line of the triangle, and also contains the triangle's de Longchamps point and nine-point center.

Easier proof

That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way.

[asy]     import olympiad;     size(4cm);     pair A=dir(110), B=dir(200), C=dir(-20), D,E,F,H;     H=orthocenter(A,B,C);     D=B+C-A;     E=C+A-B;     F=A+B-C;     draw(A--B--C--A);     draw(D--E--F--D);     draw(A--H,dotted);     draw(B--H,dotted);     draw(C--H,dotted);     label("A",A,N);     label("B",B,dir(190));     label("C",C,dir(-10));     label("D",D,S);     label("E",E,dir(45));     label("F",F,dir(140));     label("H",H,dir(50)); [/asy]

Let the line through $B$ parallel to $AC$ and the line through $C$ parallel to $AB$ intersect at $D.$ Define $E,F$ similarly.

(Alternatively, $\triangle DEF$ can be constructed by reflecting $\triangle ABC$ across each of its edges' midpoints, respectively.)

Note that $FA=BC=AE$ (and likewise for the other sides $AB$ and $CA$ of $\triangle ABC$), and so each altitude of $\triangle ABC$ is a perpendicular bisector of $\triangle DEF$.

Since the perpendicular bisectors of $\triangle DEF$ intersect (at its circumcenter), this intersection point is also the the intersection of altitudes of $\triangle ABC$, its orthocenter.


  • The orthocenter and the circumcenter of a triangle are isogonal conjugates.
  • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle.
  • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$:

[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label("$A$",A,(0,1));label("$B$",B,(-1,0));label("$C$",C,(1,0)); label("$A'$",A1,(0,-1));label("$B'$",B1,(1,1));label("$C'$",C1,(-1,1)); label("$H$",H,(-1,-1)); [/asy]

  • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle.


Art of Problem Solving Volume 2 - Example 21-4 Euclidean Geometry in Mathematical Olympiads by Evan Chen - Section 1.3

See Also