CircleHardBloom L3

Question

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:

Options

A.$\frac{AB \cdot AD}{\sqrt{AB^{2} + AD^{2}}}$
B.$\frac{AB \cdot AD}{AB + AD}$
C.$\sqrt{AB \cdot AD}$
D.$\frac{AB \cdot AD}{\sqrt{AB^{2} - AD^{2}}}$

Solution

[Key Concept: Semicircle on a leg as diameter: the angle in a semicircle is $90^{\circ}$ (use power of a point).]

$\left( AC^{2} \right) = (AD)^{2} + (CD)^{2}$

$$\begin{matrix} & \ (AB)^{2} = (BD)(BC) = BD(BD + DC) \\ \Rightarrow & \ (BD)(DC) = (AB)^{2} - (BD)^{2} = (AD)^{2} \\ \Rightarrow & DC = \frac{AD^{2}}{BD} = \frac{(AD)^{2}}{\sqrt{(AB)^{2} - (AD)^{2}}} \end{matrix}$$

$$\begin{matrix} \therefore\ (AC)^{2} & \ = (AD)^{2} + \frac{(AD)^{4}}{\left( AB^{2} \right) - (AD)^{2}} \\ & \ = \frac{(AB)^{2}(AD)^{2}}{(AB)^{2} - (AD)^{2}} \\ AC & \ = \frac{(AB)(AD)}{\sqrt{(AB)^{2} - (AD)^{2}}} \end{matrix}$$

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