The circle passes through vertices A and C of triangle ABC and intersects its sides AB and BC at points K

The circle passes through vertices A and C of triangle ABC and intersects its sides AB and BC at points K and E, respectively. Segments AE and CK are perpendicular find angle KCB if angle ABC = 20 °

The inscribed angles AKC and AEC are based on the arc AC, therefore, the angle AKC = AEC.

The BKС angle is adjacent to the AKE angle, the BEA angle is adjacent to the AEC angle, then the BKC angle = BEA.

In the quadrangle BKOE, the angle KBE = 20, the angle КОЕ = 90, then the angle BKO = BEO = (360 – 20 – 90) / 2 = 125.

Angle AEC = 180 – 125 = 55.

Angle OCE = KCB = 90 – 55 = 35.

Answer: Angle BCK is 35.