The segment BD is the diameter of the circle with the center O. Chord AC bisects the radius of the OB and is perpendicular to it. Find the angles of the quadrangle ABCD and the degree measures of the arcs AB BC CD AD.
The inscribed angles BAD and BCD are based on the arc BCD and BAD, and since BD is the diameter of the circle, the inscribed angles BAD = BCD = 180/2 = 90.
Let’s construct the radius OA.
In the triangle AOB, OA = OB = R, and the height of AH divides, by condition, BO in half, then AH is the median, which means that the triangle AOB is equilateral, and all of its internal angles are 60.
Similarly, the angle OBC = 60, then the angle ABC = 60 + 60 = 120.
Since ABCD is inscribed in a circle, the angle ADC = 180 – ABC = 180 – 120 = 60.
Then the angle ВDА = ВDC = 60/2 = 30.
Then the arc AB = BC = BDA * 3 = 30 * 2 = 60.
Arc CD = AD = 180 – 60 = 120.
Answer: The angles of the quadrilateral ABCD are equal to 120, 90, 60, 90.
Arc AB = BC = 60, arc CD = AD = 120.