Triangle AltitudesDate: 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/ |
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