On the circle with center O , points A, ~B are such that OA = AB . A point C is located on the tangent at B to the circle such that A and C are on the opposite sides of the line O B= and A B=B C . The line segment AC intersects the circle again at F . Then the ratio B O F: B O C is equal to:

Step-by-step solution

1. A O B is equilatrual ( A O B= O A B= O B A=60^ ) 2. O B C is right angled isosceles ( O B C=90^ ) 3. A B C is isosceles ( B A C= B C A=15^ . ) 4. O A C=60^ - C A B=45^ 5. A O F is right angled isosceles ( A O F=90^ , O F A=45^ ) 6. B O F=90^ - A O B=30^ 7. O B C is right angled isosceles ( B O C=45^ ) B O F B O C = 30^ 45^ = 2 3