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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
$\frac{{AB \cdot AD}}{{\sqrt {A{B^2} + A{D^2}} }}$
$\frac{{AB \cdot AD}}{{AB + AD}}$
$\sqrt {AB \cdot AD} $
$\frac{{AB \cdot AD}}{{\sqrt {A{B^2} - A{D^2}} }}$
Solution
$\therefore$ $\frac{{AC}}{{AD}}$ $= \frac{1}{{\sin \theta }}$
$\Rightarrow$ $\frac{{AD}}{{AB}}$ $=sin \,\left( {\frac{\pi }{2} – \theta } \right)$ $=cos\,\,\theta$
$\Rightarrow$ $\frac{{A{D^2}}}{{A{C^2}}} + \frac{{A{D^2}}}{{A{B^2}}}$ $=1$
$\Delta = $ $\frac{1}{2}$ $\sqrt {{x^2} + {{\rm{l}}^2}} \,\cdot\,y$ $=x^2 l^2$
$\Rightarrow$ $l=$$\frac{{xy}}{{\sqrt {{x^2} – {y^2}} }}$ $= \frac{{AB \cdot AD}}{{\sqrt {A{B^2} – A{D^2}} }}$
