Let $ABC$ and $A'B'C'$ be two non-congruent homothetic triangles whose sides are respectively parallel (Both are acute and $A'B'C'$ is inside $ABC$). How to show the three lines $AA'$, $BB'$, and $CC'$ are concurrent?
I know to prove these lines are concurrent, you must use Ceva's theorem, but I am not exactly sure how to show they all intersect.
Put both triangles onto the complex plane. Then we can assume $A = z_1$, $B = z_1 + v$. $C = z_1 + w$, $A' = z_2$, $B' = z_2 + lv$, $C' = z_2 + lw$, where $z_1, z_2, v, w \in \mathbb C$ and $l \in \mathbb R$.
To find the coordinate of the intersection of $\overline{AA'}$ and $\overline{BB'}$, note that every point on $\overline{AA'}$ have the form $z_1 + m(z_2 - z_1)$ for $m \in \mathbb R$, and every point on $\overline{BB'}$ have the form $z_1 + v + m'(z_2 + lv - z_1 - v)$, where $m' \in \mathbb R$.
So for the intersection, you need to solve for the equation $$ z_1 + m(z_2 - z_1) = z_1 + v + m'(z_2 + lv - z_1 - v), $$ or $$ (m'-m)(z_2-z_1) + \left(1 + m'(l-1)\right)v = 0. $$
If the two triangles are in general position (i.e., no edge or vertex coincide, and the two triangles are not merely parallel transports of each other), then since $z_2 - z_1$ in not a real multiple of $v$, the intersection occurs when $m'-m = 0$ and $1 + m'(l-1) = 0$.
Solve for $m$ (or $m'$, anyway they are equal). This will give you the intersection of $\overline{AA'}$ and $\overline{BB'}$. Note that anyway the intersection only depends on $l$ and $z_1, z_2$, and not on $v$, so indeed $\overline{AA'}$, $\overline{BB'}$ and $\overline{CC'}$ are concurrent.
And that should give you some insight to write down a purely geometric proof (without Ceva).