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  4. 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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Q. 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 -

Conic Sections

Solution:

Triangles $BAC$ and $BDA$ are similar
$\therefore \frac{ AC }{ AD }=\frac{ BC }{ AB }$
$A C=\frac{B C \cdot A D}{A B}$
image
$\left\{ AB ^2= BD . BC \right\}$
$=\frac{A B \cdot A D}{B D}=\frac{A B \cdot A D}{\sqrt{A B^2-A D^2}}$