# Segment BD – diameter of a circle with center O. Chord AC bisects the radius OB

**Segment BD – diameter of a circle with center O. Chord AC bisects the radius OB and is perpendicular to it. Find the angles of the quadrilateral ABCD and the degree measure of the arcs AB, BC, CD, AD.**

In triangles BCD and BAD, the angles BCD and BAD are based on the diameter BD, therefore, the angle BCD = BAD = 90.

Let’s build the radii ОC and ОD. According to the condition, the chord AC divides the radius of the ВO in half and is perpendicular to it, then CH is the height and median of the triangle BCO, and then this triangle is isosceles, BC = OC = R, and since OB = R, then the BOC is an equilateral triangle, then all of its the angles are 60. Then the angle ВDC of the triangle ВСD is 90 – 60 = 30.

Similarly, in triangle ABD, the angles are 90, 60, 30.

Then the angle ABC = 2 * 60 = 120, the angle ADC = 2 * 30 = 60.

The arc AB is equal to the central angle AOB = 60, the arc BC is also equal to 60.

The central angle AOD = COD = 180 – 30 – 30 = 120, then the arc AD = CD = 120.

Answer: The angles of the quadrangle ABCD are equal to 120, 90, 60, 90, arc AB = BC = 60, arc AD = CD = 120.