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Triangle Altitudes


Date: 03/05/99 at 15:21:28
From: Stephanie
Subject: Concurrency of the Altitudes of a Triangle

I have to prove that the three altitudes of a triangle are concurrent. 

Please help. 


Date: 03/06/99 at 03:31:18
From: Doctor Floor
Subject: Re: Concurrency of the Altitudes of a Triangle

I suppose you want to know a proof why the three altitudes in a 
triangle always have one common point of intersection ("the three 
altitudes are concurrent").

To prove this, we first prove that the three perpendicular bisectors of 
a triangle ABC are concurrent:

Recall that the perpendicular bisector of AB, passing through midpoint 
M of AB, is the set of all points that have equal distances to A and B. 
We should of course check that this is true:

1. Let P be a point on the perpendicular bisector of AB. It is easy to 
see that triangles PMB and PMA are congruent: PM = PM, MA = MB and 
angle(PMA) = angle(PMB), so we have SAS. From the congruency we can 
conclude that PA = PB. The conclusion is that all the points on the 
perpendicular bisector are at equal distances from A and B.

2. Let Q be a point such that QA = QB. Then AQB is an isosceles 
triangle. In an isosceles triangle the line through the top-angle (Q) 
perpendicular to its opposite side (AB) intersects this side in its 
midpoint (M). So Q is on the perpendicular bisector of AB. The 
conclusion is that all the points that are at equal distances from A 
and B are on the perpendicular bisector.

So, we have that the perpendicular bisector of AB is the set of all 
points D such that DA = DB. In the same way the perpendicular bisector 
of AC is the set of all points E such that EA = EC. Let P be the 
intersection point of these perpendicular bisectors; then we see that 
PA = PB = PC. So, in particular PB = PC, so P lies on the perpendicular 
bisector of BC too. And thus the three perpendicular bisectors of a 
triangle are concurrent.

But our assignment was to prove that the three altitudes in a triangle 
are concurrent. We are going to use a trick for that. In our triangle 
ABC we draw lines through A parallel to BC, through B parallel to AC, 
and through C parallel to AB. In this way we form a new triangle UVW:

          V---C---U
           \ / \ /
            A---B
             \ /
              W

In this new triangle, A is the midpoint of VW, B is the midpoint of UW, 
and C is the midpoint of VU. Now consider the perpendicular bisector of 
UV: it passes through C, and is perpendicular to UV. But if it is 
perpendicular to UV, it is also perpendicular to AB, because AB and UV 
are parallel. So the perpendicular bisector of UV is the altitude from 
C in triangle ABC.

We can conclude that the perpendicular bisectors of UVW are the 
altitudes in ABC. Since we have proven that the perpendicular bisectors 
of UVW are concurrent, the same is true for the altitudes in ABC.

The point of concurrency of the three altitudes is usually called "the 
orthocenter" and is given the letter H.

If you have a math question again, please send it to Dr. Math!

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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