Given circle $E$ with diameter $\overline{C D}$ and radius $\overline{E A} \cdot \overline{A B}$ is tangent to $E$ at $A$. If $A C=5$ and $C D=6$, solve for $A D$. Round your answer to the nearest tenth if necessary. If the answer cannot be determined, click "Cannot be determined."

Step 1 ：Given circle $E$ with diameter $\overline{C D}$ and radius $\overline{E A}$. $\overline{A B}$ is tangent to $E$ at $A$. We are given that $A C=5$ and $C D=6$, and we are asked to find $A D$.

Step 2 ：From the given information, we can infer that triangle $A C D$ is a right triangle because the radius of a circle is perpendicular to the tangent line at the point of tangency.

Step 3 ：Therefore, we can use the Pythagorean theorem to solve for $A D$. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We can write this rule as: $a² + b² = c²$.

Step 4 ：In triangle $A C D$, $A C$ is one leg, $C D$ is the other leg, and $A D$ is the hypotenuse. We can substitute the given values into the Pythagorean theorem and solve for $A D$.

Step 5 ：Substituting the given values, we get $A C = 5$, $C D = 6$, and $A D = \sqrt{A C^2 + C D^2} = \sqrt{5^2 + 6^2} = 7.810249675906654$.

Step 6 ：Final Answer: The length of $A D$ is approximately \(\boxed{7.8}\).