In a circle centered on the AC and BD diameters. The center angle of the AOD is 138 degrees.

In a circle centered on the AC and BD diameters. The center angle of the AOD is 138 degrees. Find the BCA inscribed angle.

Consider a circle centered at point O, AC and BD – the diameters of the circle that intersect at point O. Angle <AOD = 138 ° formed by the intersection of the diagonals.
To determine the angle <BCA, draw a straight line BC, we have a triangle BCA.
Consider the triangle BCO, <C = 138 °, as the vertical angle at the intersection of straight lines and opposite the angle <AOD. The sides of the triangle ВO and OС are equal, as are the radii of a circle. So the triangle is isosceles, <B = <C.
The sum of the angles of a triangle is 180 °, which means:
138 + 2 * <B = 180
<B = (180-138) / 2 = 21 °
<BCA = <C
Answer: <BCA = 21 °