An orthocenter of a triangle is the point where the lines containing its altitudes are concurrent.

A construction of an orthocenter


H the orthocenter happens to be inside the triangle. This is not always the case. If the triangle is a right triangle then H is the point where the legs intersect. If the triangle is obtuse then H will be outside the triangle. To move a verex around or play with this triangle click here to get to the GSP file that contains this sketch.

It appears that the altitudes are concurrent, this requires proof.

It can be proved that the perpidicular bisectors of a triangle are concurrent. This theorem is necessary in the proof that the altitudes of a triange are concurrent. To see a proof about the perpendicular bisectors click here.


Start with triange ABC. Construct through each vertex a line parallel to the opposite side.

Now KACB and AMCB are parallelograms by construction. So KA = BC and AM = BC since the opposite sides of a parallelogram are congruent. AD is perpendicular to BC by the definition of an altitude, therefore AD is perpendicular to KM because BC is parallel to KM.

Because KA = AM it follows that AD is the perpendicular bisector of KM a side of triangle KLM. Similiarly BE and CF are perpendicular bisectors of sides KL and LM of triangle KLM. Since AD, BE, and CF are perpendicular bisector of triangle KLM they are concurrent. Therfore the altitudes of a triangle are concurrent.