Altitudes of a Triangle

By

Janet M. Shiver

EMAT 6680

 




 

 

Investigation: Construct the three altitudes of a triangle and prove that they are concurrent.

 



 

 


We will begin our investigation by looking at several triangles and their altitudes.

Here we see an acute triangle. It appears that all three of the altitudes intersect in a single point.

Here we see an obtuse triangle. It appears that the three altitudes of the triangle once again intersect in a single point, however this time they intersect outside the triangle.

This time we will look at a right triangle.  It appears that the three altitudes are now intersecting in a single point at one of the vertices.

 

 

 

 

Necessary definitions and theorems

 

 

 

 

Definition of concurrent:

                   A number of lines are concurrent if they have exactly one point in common.

 

Definition of altitude of a triangle:

A line segment drawn from a vertex of the triangle perpendicularly to the side opposite the vertex.

 

Definition of orthocenter:

The point of concurrence for the three altitudes of a triangle.

 

Theorem of perpendicular bisectors:

The three perpendicular bisectors of the sides of a triangle are concurrent.

 

 

 

 

 

 

 

Outline of proof:

We will prove that the three altitudes of a triangle are concurrent by constructing three auxiliary lines to form a new triangle and showing that the altitudes of the of the original triangle are perpendicular bisectors of the sides of the newly formed triangle. We will then use the theorem for perpendicular bisectors to prove concurrence.

 

 

 

Proof:

 

ConsiderD ABC with altitudes  By definition of an altitude of a triangle,    is perpendicular to ,  is perpendicular to and   is perpendicular to .

                                                                                                           

 

 

 

Construct  || and passing through point A.

 

 

Similarly, construct  || and passing through point C and  || and passing through point B.

 


 

 

Since  || and  is a transversal through these segments, angle AQB is congruent to angle DAQ by the congruence of alternate interior angles. In addition, by the definition of perpendicular angle AQB is a right angle. Thus angle  DAQ must also be a right angle. Therefore,   is perpendicular to by definition of perpendicular lines.

 

 

Using a similar argument, we can show that   is perpendicular to and   is perpendicular to .

 

Next we will show that each of the altitudes is a segment bisector. By construction  || and  || . By definition of a parallelogram, ABCD forms a parallelogram and thus AD=BC.

By construction  || and  || .By definition of a parallelogram, ACBE forms a parallelogram and thus EA=BC.

Using the transitive property of equality, EA = AD.  By definition of midpoint, A is the midpoint of  and separates the segment  into two congruent parts  and  . Thus,  is the perpendicular bisector of  .

 

 

We can use a similar argument to show that  is the perpendicular bisector of  and that  is the perpendicular bisector of  .  Since we have shown that each segment  , and  is a perpendicular bisector then all three segments must be concurrent by the theorem for the concurrence of the perpendicular bisectors of a triangle.

 

 

Therefore, since the concurrent perpendicular bisectors of the sides of triangle DEF are also the altitudes of triangle ABC, these altitudes must be concurrent.

 

 

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