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Geometry Level 3

$\text{E}$ is the midpoint of line segment $\text{CD}$ of quadrilateral $\text{ABCD}$ such that $\widehat{\text{CDA}} = \widehat{\text{DAB}} = 90^\circ$ and $\displaystyle \frac{\text{BE}}{\text{BC}} = \frac{\sqrt{17}}{5}$ . If $\widehat{\text{BCD}} = 2\widehat{\text{EAD}}$ then what is the value of $\displaystyle \frac{\text{CD}}{\text{AB}}$ ?

2\sqrt{3}

\sqrt{13}

3\sqrt{2}

\sqrt{15}

\sqrt{17}

\sqrt{19}

\sqrt{14}

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