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# Circles Example

- Question:
- In the figure below, point O is the center of both small and large circles. The large circle has a radiu R = OF = 2 and small circle has a radius r = OG = sqrt(3). A is a point on the large circle, chord AB and AC are tangent to the small circle at points D and E respectively. What is the measure of arc BFC in degree ?

- Solution:
- Connect points OA, OD and OE.
- Let a = angle OAD. Since D and E are points of tangency to the small circle, so OD is perpendicular to AB and OE is perpendicular to AC (A radius drawn to a point of tangency is perpendicular to the tangent.) and also AE = AD (If two tangent segments are drawn to a circle from the same exterior point, then the two tangent segments are equal.)
- In triangle ODA and triangle OEA, since OD = OE = r, OA is in common, and AE = AD, so the triangle ADO and triangle AEO are congruent (SSS Postulate). So angle EAO = angle DAO (In congruent triangles, corresponding angles are equal.)
- Given: r = sqrt(3) and R = 2. Note: OA = R = 2.
- sin(a) = (opposite side of the angle a)/hypotenuse = OD/OA = sqrt(3)/2, then the angle a = 60
^{o}. Angle DAE = (2)(a) = 2(60^{o}) = 120^{o}. Since point D lies on AB and point E lies on AC, so angle DAO = angle BAO and angle EAO = angle CAO, so angle BAC = angle DAE = 120^{o}. - Therefore, the measure of the arc BFC = (2)(angle of BAC) = (2)(120
^{o}) = 240^{o}(The measure of an inscribed angle in a circle is equal to one-half the measure of its intercepted arc.)