2007-II-49

Join $BC$. Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACB$,

$\begin{array}{rcll}\mathrm{\angle }BAD& =& \mathrm{\angle }CAB& \text{(common angle)}\\ \mathrm{\angle }ACB& =& {90}^{\circ }& \text{(}\mathrm{\angle }\text{ in semi-circle)}\\ \mathrm{\angle }ABD& =& {90}^{\circ }& \text{(radius }\perp \text{ tangent)}\\ \therefore \mathrm{\angle }ABD& =& \mathrm{\angle }ACB\end{array}$

$\begin{array}{rcll}\mathrm{\angle }ADB& =& {180}^{\circ }–\mathrm{\angle }BAD–\mathrm{\angle }ABD& \text{(}\mathrm{\angle }\text{s sum of }\mathrm{\Delta }\text{)}\\ & =& {180}^{\circ }–\mathrm{\angle }CAB–\mathrm{\angle }ACB& \text{(proved)}\\ & =& \mathrm{\angle }ABC& \text{(}\mathrm{\angle }\text{s sum of }\mathrm{\Delta }\text{)}\end{array}$

$\therefore \mathrm{\Delta }ABD\sim \mathrm{\Delta }ACB$ (A.A.A.).

Therefore, we have

$\begin{array}{rcl}{\displaystyle \frac{AD}{AB}}& =& {\displaystyle \frac{AB}{AC}}\\ {\displaystyle \frac{6}{AB}}& =& {\displaystyle \frac{AB}{4}}\\ A{B}^2& =& 24\\ AB& =& \sqrt{24}\\ & =& 2\sqrt{6}\end{array}$