Proofs with Isosceles Triangles

Date: 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/