On Ceva's Theorem? The famous Ceva's Theorem on a triangle $\Delta \text{ABC}$


  $$\frac{AJ}{JB}  \cdot \frac{BI}{IC} \cdot \frac{CK}{EK} = 1$$

is usually  proven using the property that the area of a triangle of a given height is proportional to its base.

Is there any other proof of this theorem (using a different property)?

EDIT:  I would like if someone can use the proof of Menelaus' Theorem.

I’m sure there is a slick proof lurking in $\mathbb{C}$. This is not it.
We first prove the left implication. Place the origin $O$ at the concurrent point as figured.

Since ratios of lengths are invariant under dilations and rotations, WLOG let the line through $B$ and $K$ be the real axis and scale the triangle such that $B=1$.
As $I,J,K$ lie on the edges of $\triangle ABC$ $$ \begin{split} I&=A+t_1(1-A)\\J&=1+t_2(C-1)\\K&=C+t_3(A-C),\end{split}\tag{*}$$ for real $t_i$, from which it follows $$\tag{**}\frac{ A-I}{ I-B}\cdot \frac{ B-J}{ J-C}\cdot \frac{ C-K}{ K-A}=\frac{t_1t_2t_3}{(1-t_1)(1-t_2)(1-t_3)}.$$ Further, $$\begin{split} I&=r_1C\\J&=r_2A \\K&=r_3.\end{split}\tag{***}$$ If we equate $(*)$ and $(***)$ and solve for the real $r_i$, then we get three complex equations in $r_i,a_i,c_i, t_i$ with imaginary parts $0$. Solving these three imaginary parts for $t_i$ then gives $$\begin{split} t_1=\frac{a_1c_2-a_2c_1}{ a_1c_2-a_2c_1-c_2} &\implies \frac{1}{1-t_1}=\frac{c_2+a_2c_1-a_1c_2}{c_2}\\t_2=\frac{a_2}{a_1c_2+a_2-a_2c_1}&\implies \frac{1}{1-t_2}=\frac{a_1c_2+a_2-a_2c_1}{a_1c_2-a_2c_1}\\t_3=\frac{c_2}{c_2-a_2}&\implies \frac{1}{1-t_3}=\frac{a_2-c_2}{a_2}\end{split}$$ and plugging these into $(**)$ gives the desired result.
To prove the right implication we follow basically the same procedure, i.e. place the origin $O$ at the intersection of Ceva lines $CI$ and $AJ$, rotate and scale so that $B=1$ and then prove $K\in\mathbb{R}$ by assuming the Ceva product is $1$, hence the Ceva lines are concurrent. The same algebraic manipulations as before show $$\begin{split} \text{Im}(K)&=t_3a_2-c_2(1-t_3)\\&=t_3a_2+c_2\frac{t_1t_2t_3}{(1-t_1)(1-t_2)}\\&=t_3a_2 +c_2\frac{a_2c_1-a_1c_2}{c_2}\frac{a_2}{a_1c_2-a_2c_1}t_3\\&=0\end{split}$$ and we conclude the result.    $\qquad\square$
Not nice, but maybe you can streamline the argument somewhere.
