In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to:

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$\frac{A B . A D}{A B + A D}$

$\frac{A B . A D}{\sqrt{A B^{2} - A D^{2}}}$

$\frac{A B . A D}{\sqrt{A B^{2} + A D^{2}}}$

$\sqrt{A B . A D}$

answer is D.

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Detailed Solution

$∠ D = 90^{\circ} In △ ABC and △ DAC ∠ C is the common angle. \begin{aligned} ∠ BAC = ∠ ADC = 90^{\circ} \\ △ ABC ~ △ DAC \end{aligned} \begin{aligned} ∴ \frac{AB}{AD} = \frac{BC}{AC} = \frac{AC}{DC} \\ ∴ \frac{x}{y} = \frac{\sqrt{x^{2} + 1^{2}}}{1} = \frac{1}{\sqrt{1^{2} - y^{2}}} \end{aligned} \begin{aligned} \Rightarrow 1 \cdot x = y \sqrt{1^{2} + x^{2}} \\ \Rightarrow 1^{2} x^{2} = y^{2} (1^{2} + x^{2}) \\ \Rightarrow 1^{2} (x^{2} - y^{2}) = x^{2} y^{2} \end{aligned} \Rightarrow 1 = \frac{xy}{\sqrt{x^{2} - y^{2}}} = \frac{AB \cdot AD}{\sqrt{{AB}^{2} - {AD}^{2}}}$
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