Through point A, a tangent (B-point of tangency) and a secant are drawn, which intersects

Through point A, a tangent (B-point of tangency) and a secant are drawn, which intersects the circle at points E and F. Find EF if AB = 9, AF = 15.

Let us construct a segment BF and prove the similarity of triangles ABE and ABF.

In triangles ABE and ABF, angle A is common. The angle ABE formed by the chord and the tangent is equal to half of the arc BE, and the inscribed angle EFB is also equal to half of the arc BE on which it rests, then the angle ABE = EFB, and the triangles ABE and ABF are similar in two angles.

Then in similar triangles: AB / AE = AF / AB.

AB ^ 2 = AE * AF.

AE = AB ^ 2 / AF = 81/15 = 5.4 cm.

Then EF = AF – AE = 15 – 5.4 = 9.6 cm.

Answer: The length of the segment EF is 9.6 cm.