The midpoint functional equation

Reduction. Put $g(x)=f(x)-f(0)$. Then $g(0)=0$, $g$ is continuous, and $g$ satisfies the same midpoint equation (the constants cancel).

Halving. Taking $y=0$: $g\!\left(\frac x2\right)=\frac{g(x)}2$, i.e. $g(x)=2g\!\left(\frac x2\right)$.

Additivity. For any $x,y$,
$$g(x+y)=2g\!\left(\frac{x+y}2\right)=2\cdot\frac{g(x)+g(y)}2=g(x)+g(y).$$

Conclusion. $g$ is continuous and additive, so by Cauchy's equation $g(x)=ax$ with $a=g(1)$. Therefore $f(x)=ax+b$ with $b=f(0)$.

Remark. The equation says that the value at the midpoint is the average of the values: the graph over every interval is "flat", and that characterises straight lines among continuous functions.