Triangle ABC is isosceles with AB=BC. There is a point D on AC such that BD = DC and AB=AD.

What is the measure of angle BAC?

 

       A        This is my picture of the problem.
      / \       Though obviously distorted, it helps
     /   \      me at least, if not others, to think
    D-----B     about the problem a lot.
   /  __/  \    I added a point E such DB is parallel
  /__/      \   to EC and A, B, and E are colinear.
 C/----------E

To sum up the givens, AD=AB=BC, and DB=DC.  E is placed in a way that AC=AE.

Call the value of angle BAC "x."  Call the value of angle ADB "y."  Note that 2y+x=180 degrees because the angles of triangle ADB must sum to 180.  Because of the iscosceles triangles, angles BAC, DCB, and CBD are all equal to x.  Since angle ABC has a value of x+y, angle CBE must have a value of 180-x-y, or just plain old y.  Because of the iscosceles triangles again, CB=AB, and BE=CD=DB.  Now, by the side-angle-side theorum, we can prove triangles ABD and CBE congruent.

Using this, angle E must have a value of y.  Angle E is congruent to angle ADB, which has a value of 2x.  So, 2x=y.

Solving for x, 2(2x)+x=180, so x=36 degrees.

Edit:  This is really strange.  I was looking back at this comment, when I noticed all the angle labels were wrong, and the diagram was messed up.  I had to edit this several-months-old comment.  Either it was always like that, and no one corrected me, or some wierd computer stuff messed up the comment!

Edited on January 26, 2005, 4:54 am

Edited on January 26, 2005, 4:56 am

Posted by Tristan on 2004-09-01 17:36:08