Find the third angle, BCA.

The sum of angles in a triangle is $180^\circ$.

$$\angle BCA = 180^\circ - \angle ABC - \angle CAB$$ $$\angle BCA = 180^\circ - 80^\circ - 45^\circ$$ $$\angle BCA = 55^\circ$$

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
We want to find side $BC$ (let's call it $a$), which is opposite $\angle CAB$ ($A$).
We are given side $AC$ (let's call it $b$), which is opposite $\angle ABC$ ($B$).
So, we use the relation:

$$\frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)}$$

$$\frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)}$$ $$BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)}$$ $$BC = \frac{50 \times 0.7071}{0.9848}$$ $$BC = \frac{35.355}{0.9848}$$ $$BC \approx 35.90$$

The length of side BC is approximately 35.90.