In a triangle $\Delta ABC$, let $M,E$ be the mid points of the sides $AC$ and $AB$, respectively. Let $N$ be an arbitrary point in the segment $AM$. We denote the intersection of $EM$ and $BN$ by $Q$. The parallel to $BA$ through $N$ intercects $BM$ at $P$ and the parallel to $AQ$ through $N$ intercects $BC$ at $S$. Prove that $PS$ and $AC$ are parallel.
I found this exercise and its solution on a geometry book. To solve it, they first let $D$ to be the intersection of $PN$ and $ME$ and $R$ the intersection of $PN$ and $AQ$. Then they state that $\Delta AMP$ and $\Delta NMP$ are similar triangles, but I can't see why is that. Can anyone help me solve this? Help will be very appreciated.
First of all, I think the beginning of your problem's statement should be changed to:
Otherwise, it wouldn't match your diagram, and moreover, the statement would be false (here's a visual counterexample).
With that in mind, let $D$ be the intersection of $PN$ and $EM$, and $F$ the intersection of $NS$ and $EM$. Here's the diagram I will be using.
First, we'll prove that $A$, $Q$ and $P$ are collinear. Since $PN$ and $AB$ are parallel, then by Thales' theorem, $\frac{MP}{PB} = \frac{MN}{NA}$, and so $\frac{MP}{PB} \frac{NA}{MN} = 1$. Notice that since $E$ is the midpoint of $AB$, this implies $$\frac{MP}{PB}\cdot \frac{BE}{EA}\cdot \frac{NA}{MN} = \frac{MP}{PB}\cdot 1 \cdot \frac{NA}{MN} = \frac{MP}{PB}\cdot \frac{NA}{MN} = 1,$$ so by Ceva's theorem, $AP$, $BN$ and $ME$ are concurrent. Since $BN$ and $ME$ intersect at $Q$, then $AP$ also goes through $Q$, and so $A$, $P$, $Q$ are collinear.
On the other hand, since $PN$ and $AB$ are parallel, $$\angle MAE = \angle MND \mbox{ and } \angle MEA = \angle MDN,$$ hence by $AA$ similarity, triangles $\Delta MAE$ and $\Delta MND$ are similar. Thus, $$\frac{AE}{DN} = \frac{EM}{DM}.$$ Likewise, $\angle MEB = \angle MDP$ and $\angle MBE = \angle MPD$, so $\Delta MEB$ and $\Delta MDP$ are similar. Then, $$\frac{BE}{DP} = \frac{EM}{DM}.$$ It follows that $$\frac{AE}{DN} = \frac{BE}{DP}$$ $$\implies \frac{AE}{BE} = \frac{DN}{DP}$$ $$\implies 1 = \frac{DN}{DP},$$ and thus $DN=DP$.
Notice that $P$ lies on line $AQ$, $F$ lies on line $NS$, and $AQ$ and $NS$ are parallel; therefore, $\angle DPQ = \angle DNF$ and $\angle DQP = \angle DFN$. By $AA$ similarity, this implies that $\Delta DPQ \sim \Delta DNF$, and so $$\frac{DP}{DN} = \frac{PQ}{FN}$$ $$\implies 1 = \frac{PQ}{FN},$$ thus $FN = PQ$.
Since $AB$ and $PN$ are parallel, by Thales' theorem we have that $$\frac{AQ}{PQ} = \frac{BQ}{NQ}. \mbox{ ( I )}$$ However, because $M$ and $E$ are the midpoints of $AC$ and $AB$, respectively, then $EM$ and $BC$ are parallel, and so by Thales' theorem, $$\frac{NQ}{BQ} = \frac{FN}{SF}$$ $$\implies \frac{BQ}{NQ} = \frac{SF}{FN}. \mbox{ ( II )}$$ It follows from (I) and (II) that $$\frac{AQ}{PQ} = \frac{SF}{FN}$$ $$\implies \frac{AQ}{SF} = \frac{PQ}{FN}$$ $$\implies \frac{AQ}{SF} = 1,$$ and thus $AQ = SF$.
Therefore, $$AP = AQ + PQ = SF + FN = NS,$$ and so $AP$ and $NS$ are parallel and congruent, from which it follows that $ANSP$ is a parallelogram. Hence, $PS$ and $AN$ are parallel. Since $C$ is on line $AN$, this implies that $PS$ and $AC$ are parallel, and so we are done.