Assignment 4: Altitudes of a triangle

By Nikhat Parveen, UGA.



In this assignment we will prove the Altitude Concurrence Theorem:

"The lines of the three altitudes of a triangle are concurrent."


We will do this by using  GSP for drawing the triangles and altitudes and then by providing mathematical reasoning for the statements to be proved.


Given Triangle ABC, L1 contains the altitude from B to AC, L2 contains the Altitude from C to AB, L3 contains the altitude from A to BC.



 We want to prove that the lines L1, L2, L3 intersect at the same point, i.e.., L1, L2, L3 are concurrent. [Two or more lines are concurrent if and only if there is a single point at which they all intersect.]

Proof: Through each vertex of triangle ABC, there is a line parallel to the opposite side.

Reason: Parallel postulate. (If a point P is not on line L, there is exactly one line through P parallel to L.)



These three lines determine a triangle DEF.

Reason: Parallels to intersecting lines also intersect.



If you look closely at the figure ACBD and ACFB is a parallelogram. Why??  Because the opposite sides are parallel.

Now, AC = DB, AC = BF  Reason..... opposite sides of a parallelogram are congruent.

DB = BF by substitution.

L1 is perpendicular to BF.

Reason: In a plane, a line perpendicular to one of two parallel lines is perpendicular to the other.

Now where does it lead us....?

L1 is the perpendicular bisector of DF. why is that?? because DB = BF , and L1 is perpendicular to DF

Likewise, L2 and L3 are the perpendicular bisectors of FE and DE by the same reasoning applied to L1.

The perpendicular bisectors of the sides of triangle DEF are concurrent by the Perpendicular Bisector concurrence theorem.

Since the lines containing the altitude of triangle ABC are the perpendicular bisectors of the sides of triangle DEF, therefore the lines containing the altitudes of triangle ABC are concurrent.

**** The point of concurrency of the lines containing the altitudes is called the orthocenter of the triangle ****



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