Nice Geometry

Đào Minh Tân's solution is more direct.

Put a point G between point A and C so that DG has a length of 5. Then, we have this ratio between lengths, since triangles DGF and ECF are similar:

(x+8)/5 = 8/3

then, cross multiply:

40 = 3x+24

Subtract 24...

3x = 16

... and divide by 3.

x = 16/3

Let $\angle CAB = \alpha$ , and $\angle CFE = \beta$ ; then, it is true that $\angle FCE = 180^\circ - \alpha$ .

Firstly, applying the law of sines on $\triangle AFD$ yields: $\frac{sin \alpha}{x + 8} = \frac{sin \beta}{5}$

We can write this as:

$\frac{sin \alpha}{sin \beta} = \frac{x + 8}{5}$

Now, applying the law of sines on $\triangle EFC$ yields: $\frac{sin \beta}{3} = \frac{sin (180^\circ - \alpha)}{8}$

Noting that $sin (180^\circ - \alpha) = sin \alpha$ , we can write this as:

$\frac{sin \alpha}{sin \beta} = \frac{8}{3}$

Thus:

$\frac{x + 8}{5} = \frac{8}{3}$

Solving for $x$ gives us $x = \frac{16}{3}$ .