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Square of the tangent from an external point

In the diagram, a secant AB is drawn from the point B which lies on the tangent meeting the circle at the point C. Click and drag any of the blue points to change the shape and orientation of the diagram. What do you observe?

Consider the triangles BCD and BAC.

  1. Why is <BCD = <BAC?
  2. Given that <CBD = <ABC (common angle), we can say that the triangles BCD and BAC are similar. Why is this so?
  3. How can we use the result from (2) to prove that the square of the length of the tangent is equal to the product of the intercepts of a secant drawn from an external point?