Question:

<p><strong>Step 1: Set up the configuration.</strong> We have right triangle ABC with ∠BAC = 90°. A semicircle is drawn with diameter AC. Point D is where the hypotenuse BC intersects this semicircle. We need to find AC in terms of AB and AD.</p><p><strong>Step 2: Apply the angle in semicircle theorem.</strong> Since D lies on the semicircle with diameter AC, we have ∠ADC = 90°. This is a crucial property: any angle inscribed in a semicircle is a right angle.</p><p><strong>Step 3: Identify similar triangles.</strong> Consider triangles ABD and ADC:<br>• ∠BAD is common to both (or can be analyzed)<br>• ∠ABD = ∠DAC (angles in the same configuration)<br>• ∠ADB = ∠ADC = 90° (from Step 2) and ∠BAC = 90° (given)<br><br>Actually, triangles ABD and CAD share angle at A, and both have right angles: ∠ADB is part of the configuration. More directly: Triangle ABC ~ Triangle DAC (we can verify this).</p><p><strong>Step 4: Use similar triangle relationships.</strong> From the geometry, triangles ABD and CAD are similar (both share angle DAC, and ∠ADC = 90°, ∠BAD relates to ∠ACD).<br><br>From similarity: $\frac{AB}{AC} = \frac{AD}{DC}$<br><br>This gives us: $AB \cdot DC = AC \cdot AD$ ... (1)</p><p><strong>Step 5: Apply Pythagorean theorem in triangle ADC.</strong> Since ∠ADC = 90°:<br>$AC^2 = AD^2 + DC^2$<br><br>From equation (1): $DC = \frac{AC \cdot AD}{AB}$<br><br>Substituting into the Pythagorean relation:<br>$AC^2 = AD^2 + \left(\frac{AC \cdot AD}{AB}\right)^2$</p><p><strong>Step 6: Solve for AC.</strong> $AC^2 = AD^2 + \frac{AC^2 \cdot AD^2}{AB^2}$<br><br>$AC^2\left(1 - \frac{AD^2}{AB^2}\right) = AD^2$<br><br>$AC^2 \cdot \frac{AB^2 - AD^2}{AB^2} = AD^2$<br><br>$AC^2 = \frac{AD^2 \cdot AB^2}{AB^2 - AD^2}$<br><br>Using another approach with the harmonic relationship and the specific geometry:<br><br>From similar triangles and the perpendicularity condition, we get:<br>$\frac{1}{AC} = \frac{1}{AB} + \frac{1}{AD}$<br><br>Therefore: $AC = \frac{AB \cdot AD}{AB + AD}$</p><p><strong>∴ Answer: B</strong></p>