# exercise_Menelaus_from_Ceva.tex

\begin{exercise}
In this exercise we are going to deduce Menelaus' theorem from corollary of Ceva's theorem.
\input{./Ceva/Figure_Menelaus_from_Ceva.tex}
Let $ABC$ be a triangle and let $A',B',C'$ be three collinear points on lines $BC,AC,AB$ respectively. Let $P$ be the point of intersection of lines $BB'$ and $AA'$.
1) In triangle $ABB'$ the Cevians $AP,BC,B'C'$ are concurrent. Verify that the following equalities follow from Ceva's theorem and its corollary ...: $$\frac{BP}{PB'}\cdot\frac{B'C}{CA}\cdot\frac{AC'}{C'B}=1$$
$$\frac{BA'}{A'C}=\frac{BC'}{C'A}+\frac{BP}{PB'}$$
2) Deduce from the two equations above that $\frac{BA'}{A'C}=\frac{BC'}{C'A}(1+\frac{CA}{B'C})=\frac{BC'}{C'A}\cdot\frac{B'A}{B'C}$ or $\frac{BA'}{A'C}\cdot\frac{CB'}{B'A}\cdot\frac{AC'}{C'B}=-1$.
\end{exercise}