Ytukyg

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:

In a triangle $Delta ABC$, let $M$,$E$ be the mid points of the sides $AC$ and $AB$, respectively...

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.