Proving the converse of Ceva's theorem

Question

This is what the converse of Ceva's theorem states.

Suppose $ABC$ is a triangle, and let $AD$, $BE$, $CF$ be the three cevians such that $$\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FC}=1$$ Then $AD$, $BE$, $CF$ are concurrent.

The proof goes by assuming that all the three cevians $AD$, $BE$, $CF$ are not concurrent (say $CF$ is not concurrent), then considering another cevian $CF'$ such that it is concurrent to $AD$ and $BE$, and applying Ceva's theorem and arriving at a contradiction as $F\neq F'$ is what we have assumed.

Is there another way of proving the converse of Ceva's theorem other than this method?

Answer 1

You can do something along these lines: affine maps preserve the ratios of areas and the ratios of lengths of segments on the same line, so you may assume without loss of generality that $\widehat{BAC}=90^\circ$ and $AB=AC=1$. Now we may describe $A$ as $(0;0)$, $F$ as $(x;0)$ and $E$ as $(0;y)$, so $P=BE\cap CF$ lies at $$ P=\left(\frac{x(1-y)}{1-xy};\frac{y(1-x)}{1-xy}\right) $$ and $Q=AP\cap BC$ lies at $$ Q=\left(\frac{x(1-y)}{x+y-2xy};\frac{y(1-x)}{x+y-2xy}\right)$$ i.e. at the only point of $BC$ such that $\frac{BQ}{QC}=\frac{EA}{CE}\cdot\frac{FB}{AF}$. This proves simultaneously Ceva's theorem and its converse.