Proving Menelaus' theorem in the complex plane $\triangle{ABC}$
$A'\in BC,~B'\in CA,~C'\in AB$
$\angle A'B'C'=\pi\Rightarrow \frac{\vec{A'B}}{\vec{A'C}}\cdot\frac{\vec{B'C}}{\vec{B'A}}\cdot\frac{\vec{C'A}}{\vec{C'B}}=1$
I need to prove in the complex plane.
My approach:
Consider x-Axis on the line $A'B'C'$ and y-Axis passing through A (to simplify things).
$A(a=y_1i),~B(b=x_2+y_2i),~C(c=x_3+y_3i),~A'(a'),~B'(b'),~C'(c'),~y_1,x_2,y_2,x_3,y_3,a',b',c'\in\mathbb{R}$
Get the conditions of collinearity for $A',~B',~C'$:
$A'\in BC\Rightarrow\frac{z_B-z_{A'}}{z_C-z_{A'}}=k_1\in\mathbb{R}\Rightarrow k_1=\frac{x_2-a'}{x_3-a'}=\frac{y_2}{y_3}\Rightarrow a'=\frac{x_2y_3-y_2x_3}{y_3-y_2}$
$B'\in AC\Rightarrow\frac{z_A-z_{B'}}{z_C-z_{B'}}=k_2\in\mathbb{R}\Rightarrow k_2=\frac{-b'}{x_3-b'}=\frac{y_1}{y_3}\Rightarrow b'=\frac{x_3y_1}{y_1-y_3}$
$C'\in AB\Rightarrow \frac{z_A-z_{C'}}{z_B-z_{C'}}=k_3\in\mathbb{R}\Rightarrow k_3=\frac{-c'}{x_2-c'}=\frac{y_1}{y_2}\Rightarrow c'=\frac{x_2y_1}{y_1-y_2}$
The equation that needs to be proved has the complex equivalent:
$\frac{b-a'}{c-a'}\cdot\frac{c-b'}{a-b'}\cdot\frac{a-c'}{b-c'}=1$
Substitute $a,~b,~c$ with the carthesian form and $a',~b',~c'$. For each fraction the denominators will simplify after amplifying $a,~b,~c$ and I'm left with a huge thing. How do I handle it? Also, am I in the right direction?
Thanks in advance!
 A: Here is an analytic solution for Menelaus' Theorem along the lines in the wish in the OP. In general, i will denote for a letter $X$, which stays for a point in the plane, by the lower case version $x$ of it the affix in $\Bbb C$, so $x$ is a complex number.
We start with the condition that $A'$, $B'$, $C'$ are on a line, and want to show the algebraic identity:
$$
\tag{$*$}
\prod\frac{a'-b}{a'-c}
:=
\frac{a'-b}{a'-c}\cdot
\frac{b'-c}{b'-a}\cdot
\frac{c'-a}{c'-b}
=1\ .
$$
(We will also use below products, in the sense of building cyclicly corresponding factors.)
First of all, note that the above relation is compatible with translations of the plane. (A difference like $a'-b$ is invariated by a translation $Tz:= z+K$, i.e. $Ta'-Tb=a'-b$.) So we may and do assume that the line $A'B'C'$ passes through the origin. The equation of such a line is $mz+n\bar z=0$, but for the sake of simplicity and typing, let us assume below $n\ne 0$, so we can norm $n=-1$ and the equation is
$$
(A'B'C')\qquad \bar z = mz\ .
$$
(One of $m,n$ has to be nonzero. In the case $n=0$ just use the complex conjugation, which invariates the realtion $(*)$ to be shown.)
We compute the affix for the intersection point $A'$ of the lines $(A'B'C')$ and $(BC)$. The latter one has the equation:
$$
(BC)
\qquad
0=
\begin{vmatrix}
1&b&\bar b\\
1&c&\bar c\\
1&z&\bar z\\
\end{vmatrix}
\ .
$$
We build the linear system with two equations, the above one and $\bar z=mz$, and solve it by inserting this expression in the right bottom corner inside the determinant. We get the equation in $z$
$$
\begin{aligned}
0&=
\begin{vmatrix}
b&\bar b\\
c&\bar c
\end{vmatrix}
-
z
\begin{vmatrix}
1&\bar b\\
1&\bar c
\end{vmatrix}
+
mz
\begin{vmatrix}
1&b\\
1&c
\end{vmatrix}
\ .
\\[3mm]
&\qquad\text{ This gives the solution $a'$ in explicit form:}
\\
a'&=\frac{b\bar c-\bar bc}{m(b-c)-(\bar b-\bar c)}\ .
\\[3mm]
&\qquad\text{ Then:}
\\
a'-b &=
\frac 1{\dots}\Big[\ \color{green}{b\bar c}-\bar bc\qquad -mb(b-c)+b(\bar b-\color{green}{\bar c})\ \Big]\qquad\text{ (green terms disappear...)}
\\
&=
\frac 1{\dots}\Big[\ \bar b(b-c)-mb(b-c)\ \Big]\qquad\text{ (and we can factor...)}
\\
&=
\frac 1{\dots}(\bar b-mb)(b-c) \ .\qquad\text{ Similarly:}
\\
a'-c&=
\frac 1{\dots}(\bar c-mc)(b-c) \ .\qquad\text{ This gives:}
\\[3mm]
\frac{a'-b}{a'-c}
&=\frac{\bar b-mb}{\bar c-mc}\ ,\qquad\text{ leading to:}
\\
\prod\frac{a'-b}{a'-c}
&=
\prod\frac{\bar b-mb}{\bar c-mc}
\\
&=
\frac{\bar b-mb}{\bar c-mc}\cdot
\frac{\bar c-mc}{\bar a-ma}\cdot
\frac{\bar a-ma}{\bar b-mb}
\\
&=1\ .
\end{aligned}
$$
$\square$

Note: I hope it is clear which are the factors cancelling above in the version of the proof written with a general equation $mz+n\bar z=0$ (or even $mz+n\bar z+p=0$) instead of our simplified version $\bar z-mz=0$.
