Solving Geometric Problem - Find \angle ATB

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SUMMARY

The discussion focuses on solving a geometric problem involving an isosceles triangle ABC with an angle A measuring 48 degrees. The angle A is bisected at point T along line segment CB, leading to the determination of angle ATB. Two solutions are established: angle ATB equals 108 degrees when angle A is one of the side angles, and 90 degrees when angle A is the top angle. The third solution is derived from the realization that the configuration of angles affects the equality of the resulting triangles.

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danne89
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Hi! My problem sounds as following:
In the isoscele triangle [itex]\triangle ABC[/itex], [itex]\angle A[/itex] = 48 degrees. Bisect the angle [itex]\angle A[/itex] against [itex]\overline CB[/itex] in point T. Determine the [itex]\angle ATB[/itex] 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 [itex]\angle ATB[/itex] = 108 degrees.

The second one you find if you select the top angle and then get [itex]\angle ATB[/itex] = 90 degrees.
 
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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?
 
Ohh, I see now. Thanks! I was being misslead by myself.
 

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