Question:

Step 1: Identify geometric properties and establish perpendicularity. We are given a triangle $ABC$ that is right-angled at $A$. A semicircle is described on the leg $AC$ as diameter. The point $D$ is the intersection of the hypotenuse $BC$ and the semicircle. Since $AC$ is the diameter of the semicircle and $D$ is a point on the semicircle, the angle subtended by the diameter at $D$ is $90^\circ$. Thus, $\angle ADC = 90^\circ$. This implies that $AD$ is perpendicular to $BC$. Step 2: Apply Pythagorean theorem and altitude relations. Since $\angle ADC = 90^\circ$, we can apply the Pythagorean theorem in the right-angled triangle $\triangle ADC$: $$AC^2 = AD^2 + CD^2 \quad \text{(Equation 1)}$$ In the right-angled triangle $\triangle ABC$, $AD$ is the altitude to the hypotenuse $BC$. From the geometric mean theorem for altitudes (which arises from similar triangles $\triangle DBA \sim \triangle DAC$), we have: $$AD^2 = BD \cdot CD \quad \text{(Equation 2)}$$ Also, from the leg rule for right triangles (or from similarity $\triangle DBA \sim \triangle ABC$), we have: $$AB^2 = BD \cdot BC \quad \text{(Equation 3)}$$ Note that the hypotenuse $BC$ can be written as the sum of its segments $BD$ and $CD$: $BC = BD + CD$. Step 3: Express $CD$ in terms of $AD$ and $AB$. From Equation 2, we can express $BD$ as: $$BD = \frac{AD^2}{CD}$$ Now, substitute this expression for $BD$ into Equation 3, replacing $BC$ with $(BD + CD)$: $$AB^2 = BD (BD + CD)$$ $$AB^2 = \left(\frac{AD^2}{CD}\right) \left(\frac{AD^2}{CD} + CD\right)$$ $$AB^2 = \frac{AD^2}{CD} \left(\frac{AD^2 + CD^2}{CD}\right)$$ $$AB^2 = \frac{AD^2 (AD^2 + CD^2)}{CD^2}$$ Multiply both sides by $CD^2$: $$AB^2 CD^2 = AD^4 + AD^2 CD^2$$ Rearrange the terms to solve for $CD^2$: $$AB^2 CD^2 - AD^2 CD^2 = AD^4$$ $$CD^2 (AB^2 - AD^2) = AD^4$$ $$CD^2 = \frac{AD^4}{AB^2 - AD^2}$$ Taking the square root of both sides to find $CD$: $$CD = \frac{AD^2}{\sqrt{AB^2 - AD^2}}$$ Step 4: Substitute the expression for $CD$ into the Pythagorean relation for $AC$ and simplify. Now, substitute the expression for $CD$ from Step 3 into Equation 1: $$AC^2 = AD^2 + \left(\frac{AD^2}{\sqrt{AB^2 - AD^2}}\right)^2$$ $$AC^2 = AD^2 + \frac{(AD^2)^2}{(\sqrt{AB^2 - AD^2})^2}$$ $$AC^2 = AD^2 + \frac{AD^4}{AB^2 - AD^2}$$ To combine the terms on the right-hand side, find a common denominator: $$AC^2 = \frac{AD^2 (AB^2 - AD^2)}{AB^2 - AD^2} + \frac{AD^4}{AB^2 - AD^2}$$ $$AC^2 = \frac{AD^2 AB^2 - AD^4 + AD^4}{AB^2 - AD^2}$$ $$AC^2 = \frac{AD^2 AB^2}{AB^2 - AD^2}$$ Finally, take the square root of both sides to find $AC$: $$AC = \sqrt{\frac{AD^2 AB^2}{AB^2 - AD^2}}$$ $$AC = \frac{\sqrt{AD^2 AB^2}}{\sqrt{AB^2 - AD^2}}$$ $$AC = \frac{AD \cdot AB}{\sqrt{AB^2 - AD^2}}$$ Step 5: Conclude the final answer. The length of $AC$ is $\frac{AB \cdot AD}{\sqrt{AB^2 - AD^2}}$. This matches Option 4. The final answer is $\boxed{\text{Option 4}}$.