I am re-reading my old first semester Physics textbook and came across the well known formula that for a particle in constant acceleration $a_x$ from time $0$ to time $t$ $$v_{av-x}=\frac{1}{2}\left(v_{0x}+v_x\right)$$ where $v_{av-x}$ is the average velocity of the particle over the interval from $0$ to $t$, $v_{0x}$ is the instantaneous initial velocity, and $v_x$ is the instantaneous velocity at time $t$. It's important to note that $v_{av-x}$ is defined as $\Delta x/\Delta t$, where $\Delta x$ is the displacement of the particle.
Clearly this can be justified with calculus using the formula $v_{av-x}=\frac{1}{t-0}\int_0^t v_x(t)\space dt$, but it seems one should be able to prove this for the constant acceleration case using algebra only, no? Of course, I thought about the "area under the $v_xt$ graph" as a way to prove this, but that's calculus in disguise.
Is there any neat way to prove this without calculus or resorting to the handwaivy "average of a linearly varying quantity" argument?
For constant acceleration $a_x$, the displacement $x(t)$ and the instantaneous velocity $v_x(t)$ are given by \begin{align} x(t)&=x_0+v_{0x}t+\frac{1}{2}a_xt^2, \\ v_x(t)&=v_{0x}+a_xt. \end{align} Therefore, using the definition of $v_{av-x}$, one has $$ v_{av-x}=\frac{\Delta x}{\Delta t}=\frac{x(t)-x(0)}{t-0}=v_{0x}+\frac{1}{2}a_xt= \frac{v_{0x}+v_x(t)}{2}. $$