What is the orthocenter? Determining the orthocenter in a triangle

The orthocenter of a triangle is the intersection of the three altitudes , that is, the intersection of the lines from each vertex of the triangle to its opposite side forming a right angle. The length of the altitude is the distance between the vertex and the base.

Orthocenter of a triangle

• What is the orthocenter?
• How to determine the orthocenter of a triangle
• Properties of the orthocenter of a triangle
• Exercises to determine and prove the orthocenter of a triangle

What is the orthocenter?

The three altitudes of a triangle meet at a point. That point is called the orthocenter of the triangle .

Specifically: In the drawing are the heights, the orthocenter of the triangle.

How to determine the orthocenter of a triangle

To determine the orthocenter of a triangle, we find the intersection of the two altitudes in that triangle.

Note:  a) If the triangle is an acute triangle, the orthocenter lies inside the triangle.

b) If the triangle is a right triangle at then the orthocenter coincides with the point .

c) If a triangle is an obtuse triangle, then the orthocenter lies outside the triangle.

Properties of the orthocenter of a triangle

Property 1:  In an equilateral triangle, the centroid, orthocenter, a point equidistant from the three vertices of the triangle, a point inside the triangle and equidistant from the three sides of the triangle are four coincident points.

Property 2:  The orthocenter cuts the perpendicular bisector of two sides into two segments of equal length. This means that the orthocenter is equal distance from the vertices of the triangle.

Property 3:  The orthocenter is the center of the circumcircle of a triangle, meaning that if we draw a circle passing through the three vertices of a triangle, the orthocenter will be the center of that circle.

Property 4:  The orthocenter of an acute triangle lies inside the triangle, while the orthocenter of an obtuse triangle lies outside the triangle.

Property 5:  The orthocenter of a right triangle coincides with the vertex of the right angle of that right triangle.

Property 6:  The orthocenter is the only point in a triangle such that if we draw lines from the orthocenter to the vertices of the triangle, the sum of the lengths of those lines is the smallest. This means that the orthocenter is closest to the vertices of the triangle than any other point.

Property 7: The orthocenter is also the center of the circumcircle of the triangle, that is, the largest circle that can be drawn through the three vertices of the triangle.

Exercises to determine and prove the orthocenter of a triangle

For example: Given a non-square triangle. Call it its orthocenter. Point out the altitudes of the triangle. From there, point out the orthocenter of the triangle.

Solution guide

Illustration

Let be the feet of the perpendiculars drawn from ΔABC.

Consider ΔHBC with:

so AD is the height from H to BC.

at F so BA is the altitude from B to HC

at E so CA is the height from C to HB.

intersect at A so A is the orthocenter of ΔHCB.

For example: Given a right triangle at with height . Let the midpoint of be , the midpoint of is . Determine the orthocenter of the triangle .

Solution guide

Consider the sub-problem if the triangle has and AC as the midpoints respectively then and .

Indeed, on the opposite ray of the ray take a point such that

Consider triangle AMN and triangle CPN.

(opposite)

, (two sides and two corresponding angles)

Two angles are in alternate positions so

=>(two alternate interior angles)

Consider triangle BMC and triangle PCM.

(cmt)

MC is a common edge

, (corresponding sides and angles)

Two angles are in alternate positions so

We have again

Consider triangle HAB with:

(as proven above)

Consider triangle ADE.

on the other hand and

is the height of triangle ADE

C is the intersection of AC and DC

=> C is the orthocenter of triangle ADE

For example: Given a triangle at A, the altitude intersects the median at . Prove and calculate?

Instruct

Illustration

Because the balance is at A and AM is the median

⇒ AM is also the altitude corresponding to BC

at M.

On the other hand, and so K is the orthocenter.

Therefore, K belongs to the altitude from C of ∆ABC.

We have: