How do I find the values for $a_0, a_1, \ldots, a_n$ such that the triangle is divided into $n+2$ parts of equal area? In the above example, $n=2$.
Let's assume that $a_i = 0$ means the line is at the lower right corner, and $a_i = 1$ is the vertical edge of the triangle. (In the above example, $a_0 \approx 0.25$, $a_1 \approx 0.5$, and $a_2 \approx 0.8$. That's obviously not a solution to the problem.)
The areas of the triangles to the right of the lines are proportional to $a_k^2$, so you want $a_{k+1}^2-a_k^2$ to be constant; that is, $a_k^2=ck+d$. The constants $c$ and $d$ are fixed by $a_{-1}=0$ and $a_{n+1}=1$, which yield $-c+d=0$, i.e. $c=d$, and $c(n+1)+d=1$, so $c(n+2)=1$ and $c=d=1/(n+2)$.
Thus $a_k=\sqrt{(k+1)/(n+2)}$.