We will use with unoriented segments. According to Menelaus' theorem, applied in triangle $ PED$ for the transversal $ \overline{AMC}$, \[ \frac {PA}{AE} \cdot \frac {EC}{CD} \cdot \frac {DM}{MP} = 1, \] and since the quadruple $ (E, B, D, C)$ is harmonic, $ EC/CD = EB/BD$. Therefore, \[ \frac {PA}{AE} \cdot \frac {EB}{BD} \cdot \frac {DM}{MP} = 1, \] and thus, by Ceva's theorem, we conclude that the lines $ EM$, $ DA$, and $ PB$ are concurrent at $ Q$ (i.e. $ PQ$ passes through $ B$).

What means Q1 ?! This problem is very easy ! anonymous1173 wrote: Consider in a triangle $ ABC$ two fixed points $ \{D,E\}\subset BC$ so that $ D\in (BC)$ and $ B\in (DE)$ . For a mobile point $ P\in AE$ such that $ A\in (EP)$ define the points $ \{\begin{array}{c} M\in DP\cap AC \\ \ Q\in ME\cap AD\end{array}$ . Prove that $ PQ$ passes through a fixed point. Proof. Denote $ F\in PQ\cap BC$ . Since the lines$ PF$ , $ AD$ , $ EM$ are concurrently ( in a point $ Q$ ), then the point $ F$ is conjugate of the point $ C\in BC\cap AM$ w.r.t the points $ D$ , $ E$ , i.e. the point $ F$ is fixed. Another particular cases. 1. $ E\in AA\cap BC$ and $ D$ - the foot of the $ A$ - symmedian $ \Longrightarrow$ $ F = B$ . 2. $ AP\parallel ME\parallel BC$ and $ D$ - the middlepoint of the side $ [BC]$ $ \implies$ $ F = B$ .

Virgil Nicula wrote: What means Q1 ?! This problem is very easy ! Q1 means Question 1 (first problem in the test). The easiness of the problem is relative to its solvers.

Valentin Vornicu wrote: The easiness of the problem is relative to its solvers. ... and to ... ? !

P can move on line AE ,the problem is also true.

it's nice!