# Geometric problem

1. Jan 6, 2005

### danne89

Hi! My problem sounds as following:
In the isoscele triangle $\triangle ABC$, $\angle A$ = 48 degrees. Bisect the angle $\angle A$ against $\overline CB$ in point T. Determine the $\angle ATB$ for the three possible solutions.

I've found two of them, but cannot find the last.

The first one is if you pick A as one of the conjugent angles and then you get $\angle ATB$ = 108 degrees.

The second one you find if you select the top angle and then get $\angle ATB$ = 90 degrees.

2. Jan 6, 2005

### BobG

If you bissect the angle against CB at point T you now have two triangles.

If A is the 'top' angle (i.e. B and C are equal), then both of the new triangles are identical to each other - it doesn't matter which of the remaining angles is B and which is C.

If A is one of the side angles (equal to either B or C), then the two new triangles are not equal to each other. If B is equal to A, then ATB is equal to 108 degrees as you said. But how about if A is equal to C and B is the top angle?

3. Jan 6, 2005

### danne89

Ohh, I see now. Thanks! I was being misslead by myself.