Proofs with Isosceles TrianglesDate: 10/28/98 at 20:55:18 From: Molly Subject: Proofs with Isosceles Triangles I don't understand the difference between angle bisectors, medians, and altitudes. Here's a problem that I have to prove: In an isosceles triangle, the altitude is a median and an angle bisector. From what I understand I know that the BD is an altitude of Triangle ABC, but I need to prove why it's also a median and an angle bisector. This is where I get really lost. Please help. Date: 10/29/98 at 18:10:25 From: Doctor Rick Subject: Re: Proofs with Isosceles Triangles Hi, Molly. To understand what facts we are given and what we need to prove, let's review some definitions. The altitude, median, and angle bisector of a triangle are all line segments that join one vertex of a triangle to the opposite side. (In some cases the side may have to be extended beyond a vertex.) -- The altitude is perpendicular to that opposite side. -- The median meets the opposite side at the midpoint of the side. -- The angle bisector bisects the angle of the triangle at the vertex. B ^ /|\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ /_________|_________\ A D C So you know that BD is perpendicular to AC. You also know, because triangle ABC is isosceles, that AB and BC are equal (and by a theorem, angles BAC and BCA are equal). You need to prove that (1) AD = DC (so BD is a median), and (2) angle ABD = angle DBC (so BD is an angle bisector). Can you do the proof now that we have spelled out what you start with and what you need to prove? Try it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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