Point E is equidistant from the vertices of triangle ABC and is located in the plane of this triangle

Point E is equidistant from the vertices of triangle ABC and is located in the plane of this triangle ACE = 32 degrees. Find the angle ABC

Since the segments AE = BE = CE, the point E is the center of the ABC described about the triangle.

The AEC triangle is isosceles, since AE = CE = R, then the angle CAE = ACE = 32, then the angle AEC = (180 – 32 – 32) = 116.

The central angle AEC rests on the arc AC, the degree measure of which is 116. The inscribed angle ABC also rests on the arc AC, then the angle ABC = 116/2 = 58.

Answer: Angle ABC is 58.