+77 In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO = 5$, and angle ABO = arc CD = 60 degrees. Find the length of $BC$.
+129850 If arc CD = 60°, the angle CAD is 1/2 of this since it intercepts this arc
So CAD = 30° = BAO
And since BAO and ABO = 30 + 60 = 90°
Then,in triangle BAO, angle BAO must = 90° = angle BOD
So....triangle BAO is a 30-60-90 right triangle
Now....draw OC....and it will be equal to OD since both are radii
And if arc CD = 60°....then angle central angle COD = 60°
And since angle BOD = 90 °
Then angle BOC = angle BOD - angle COD = 90 - 60 = 30°
And angle ABO = 60 so angle CBO is supplemental = 120°
So angle BCO = 180 - angle BOC - angle CBO = 180 - 30 - 120 = 30°
Thus triangle BOC is isosceles with angle BOC = angle BCO
But BO = 5....so.....BC has the same measure = 5
