<s>[INST] 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.
 [/INST] The following proof is heavily influenced by my background in projective geometry. Use homogenous and choose the following affine basis, without loss of generality. You might also consider this as barycentric coordinates, the way Olivier Bégassat wrote in a comment.
\begin{align*}
 A &= \begin{pmatrix}1\\0\\0\end{pmatrix} &
 B &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
 C &= \begin{pmatrix}0\\0\\1\end{pmatrix}
\end{align*}
Based on these coordinates, you can specify the other three points as linear combinations of the corresponding triangle corners:
\begin{align*}
 I &= \begin{pmatrix}0\\\lambda_I\\\mu_I\end{pmatrix} &
 J &= \begin{pmatrix}\lambda_J\\\mu_J\\0\end{pmatrix} &
 K &= \begin{pmatrix}\mu_K\\0\\\lambda_K\end{pmatrix}
\end{align*}
You can obtain the oriented length ratios from these parameters:
\begin{align*}
\frac{AJ}{JB} &= \frac{\mu_J}{\lambda_J} &
\frac{BI}{IC} &= \frac{\mu_I}{\lambda_I} &
\frac{CK}{KA} &= \frac{\mu_K}{\lambda_K}
\end{align*}
Now you can compute the connections of each of these points with the opposite triangle corner using a cross product:
\begin{align*}
 A\times I &= \begin{pmatrix}0\\-\mu_I\\\lambda_I\end{pmatrix} &
 B\times K &= \begin{pmatrix}\lambda_K\\0\\-\mu_K\end{pmatrix} &
 C\times J &= \begin{pmatrix}-\mu_J\\\lambda_J\\0\end{pmatrix} \\
\end{align*}
To check whether these three lines are concurrent, you compute their determinant. If that determinant becomes zero, the lines go through the same point.
\begin{align*}
\begin{vmatrix}
0 & \lambda_K & -\mu_J \\
-\mu_I & 0 & \lambda_J \\
\lambda_I & -\mu_J & 0
\end{vmatrix} =
\lambda_I\lambda_J\lambda_K - \mu_I\mu_J\mu_K &= 0 \\
\lambda_I\lambda_J\lambda_K &= \mu_I\mu_J\mu_K \\
1 &= \frac{\lambda_I\lambda_J\lambda_K}{\mu_I\mu_J\mu_K} =
\frac{AJ}{JB}\cdot\frac{BI}{IC}\cdot\frac{CK}{KA}
\end{align*}
As the choice of coordinates was without loss of generality, the above equivalence (between the concurrence of the three lines and the product of oriented length ratios being one) will hold for any non-degenerate triangle $ABC$.
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