Don't Trust Your Eyes

Assume that : $CE = x$ , $DE = y$

That's mean $CD = x-y$

Using Pythagoras :

$BC = \sqrt{900 - x^2}$

$EG = \sqrt{1600 - x^2}$

Using Similiar Triangles :

$\LARGE \frac{FD}{BC} = \frac{DE}{CE} \\ \LARGE \frac{15}{\sqrt{900-x^2}} = \frac{y}{x} \\ \LARGE y =\frac{15x}{\sqrt{900-x^2}}$

$\LARGE \frac{FD}{EG} = \frac{CD}{CE} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-y}{x} \\ \LARGE \frac{15}{\sqrt{1600-x^2}} = \frac{x-\frac{15x}{\sqrt{900-x^2}}}{x}$

$\LARGE 15x = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}})$

$\LARGE 0 = \sqrt{1600-x^2} \times ( x - \frac{15x}{\sqrt{900-x^2}}) - 15x$

$\LARGE x = 15.9876...$

$\LARGE x \approx \color{#D61F06}{\boxed{16}}$

$\color{#D61F06}{\spadesuit} \color{#20A900}{\clubsuit}$ Thank you for @Abhay Kumar @Mark C @Ashish Siva and others for solving this problem $\color{#D61F06}{\clubsuit} \color{#20A900}{\spadesuit}$