The correct answer is **B) $\left( \sqrt{58} - 2 \right)$ units** **Step-by-step Solution:** 1. **Identify the geometric configuration and relevant theorems:** We are given a triangle $\triangle ABC$ with its angle bisector $AD$ intersecting the circumcircle at $E$. This configuration involves properties related to angle bisectors and cyclic quadrilaterals (or similar triangles formed by chords). 2. **Recall the property relating angle bisector, circumcircle, and sides:** When an angle bisector $AD$ of $\triangle ABC$ intersects the circumcircle at $E$, there is a well-known property derived from similar triangles: $AB \cdot AC = AD \cdot AE$. Let's quickly derive this property for clarity: Consider $\triangle ABD$ and $\triangle AEC$. * $\angle BAD = \angle CAE$ (Since $AD$ is the angle bisector of $\angle A$, so $\angle BAD = \angle CAD$. $E$ lies on the extension of $AD$, so $\angle CAE$ is the same as $\angle CAD$). * $\angle ABD = \angle AEC$ (Angles subtended by the same arc $AC$ in the circumcircle are equal). By Angle-Angle (AA) similarity criterion, $\triangle ABD \sim \triangle AEC$. From the similarity, the ratio of corresponding sides must be equal: $\frac{AB}{AE} = \frac{AD}{AC} = \frac{BD}{EC}$ From the first equality, $\frac{AB}{AE} = \frac{AD}{AC}$, we can cross-multiply to get: $AB \cdot AC = AD \cdot AE$ 3. **Substitute the given values into the equation:** Let $AD = x$ units. We are given: * $AB = 6$ units * $AC = 9$ units * $DE = 4$ units Since $E$ lies on the extension of $AD$, the length $AE$ can be expressed as $AD + DE$. So, $AE = x + 4$ units. Substitute these values into the property $AB \cdot AC = AD \cdot AE$: $6 \cdot 9 = x(x + 4)$ $54 = x^2 + 4x$ 4. **Solve the resulting quadratic equation for $x$ (the length of AD):** Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$: $x^2 + 4x - 54 = 0$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Here, $a=1$, $b=4$, $c=-54$. $x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-54)}}{2(1)}$ $x = \frac{-4 \pm \sqrt{16 + 216}}{2}$ $x = \frac{-4 \pm \sqrt{232}}{2}$ 5. **Simplify the square root and find the valid length:** Simplify $\sqrt{232}$: $\sqrt{232} = \sqrt{4 \cdot 58} = 2\sqrt{58}$ Substitute this back into the expression for $x$: $x = \frac{-4 \pm 2\sqrt{58}}{2}$ $x = -2 \pm \sqrt{58}$ Since $AD$ represents a length, it must be a positive value. We have two possible solutions: * $x_1 = -2 + \sqrt{58}$ * $x_2 = -2 - \sqrt{58}$ As $\sqrt{58}$ is approximately $7.6$ (since $7^2 = 49$ and $8^2 = 64$), $x_1 = -2 + 7.6 = 5.6$ (approximately), which is a positive value. $x_2 = -2 - 7.6 = -9.6$ (approximately), which is negative and thus not a valid length. Therefore, the length of $AD$ is $\left( \sqrt{58} - 2 \right)$ units. The final answer is $\boxed{\left( \sqrt{58} - 2 \right) \text{ units}}$