Multiple choice

On the circle with centre O, points A and 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 lines OB and AB$=$BC. The line segment AC intersects the circle again at F. Then the ratio $\angle$BOF$:\angle$BOC is equal to:

  1. $1:2$
  2. $2:3$
  3. $3:4$
  4. $4:5$
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B Correct answer
Explanation

Using geometric properties of the circle and tangent, one can derive the angles. The configuration leads to specific triangle relationships where the ratio of the angles BOF and BOC is 2:3.

AI explanation

Since OA equals AB in triangle OAB, it is an isosceles triangle with angle AOB = angle ABO = 60 degrees, making angle OAC = 120 degrees. Given AB = BC, triangle ABC is also isosceles, meaning angle BCA = angle BAC = 30 degrees, so the inscribed angle FAC equals 30 degrees. By the Inscribed Angle Theorem, the central angle BOF is twice the inscribed angle FAC, giving angle BOF = 60 degrees. Since OC is perpendicular to the tangent at B and angle OBC is 90 degrees, the exterior angle BOC of triangle ABC evaluates to 90 degrees, making the ratio of angle BOF to angle BOC equal to 2:3.

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