3) DH=9.

The diagram shows a secant DFE and a tangent DH. The lengths are DE = 6, DF = x, and DH = 9. According to the tangent-secant theorem, the square of the length of the tangent segment from an external point to a circle is equal to the product of the lengths of the secant segment from the external point to the farther intersection point and the external secant segment from the external point to the nearer intersection point. Thus, DH^2 = DE * DF. Substituting the given values, we have 9^2 = 6 * x. This simplifies to 81 = 6x. Solving for x, we get x = 81 / 6 = 13.5.