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?

I read through some of these, and at first, I came up with 45 degrees, because that's the way I drew my triangle.

I then got to wondering and intentionally drew a very "sqatty" iso triangle. I made it a height of 2 and equal sides of 5. From the calculations it came out that the base AC was 9.16, and anglea A and C were 23.5 degrees.

Now, draw a line BD such that angle DBC is 23.5 degrees and you will find that BD is equal to CD, and is basically just a smaller version of the original triangle.

So the answer I have come up with is ANY angle between 1 and 59 degrees will work to satisfy the givens

Posted by Jim on 2004-09-13 14:38:35