Let M be an interior point of the parallelogram ABCD. The parallel lines through M to the sides of the parallelogram intersect AB, BC, CD, and DA at Q, P, R, and N, respectively. Let point S be defined as the intersection of BN and DQ. Prove that points C, M, and S are collinear. [![enter image description here](https://i.stack.imgur.com/l6nJ2.png)](https://i.stack.imgur.com/l6nJ2.png)

So far I am trying to use similar triangles. I have that:

$\triangle DNS \sim \triangle DAQ$ by AA so $\displaystyle\frac{DN}{DA} = \frac{DS}{DQ} = \frac{NS}{AQ}$ 

$\triangle DSN \sim \triangle BSQ$ so $\displaystyle\frac{BS}{NS} = \frac{QS}{DS} = \frac{QB}{DN}$ 

$\triangle QSM \sim \triangle QDR$ so $\displaystyle\frac{QS}{QD} = \frac{QM}{QR} = \frac{SM}{DR}$ 

$\triangle NSM \sim \triangle NBP$ so $\displaystyle\frac{NS}{NB} = \frac{NM}{NP} = \frac{MS}{BP}$

I know to prove $C$, $M$, and $S$ collinear, to use Menelaus' C-M-S is the transversal. The two triangles I found with this transversal are $\triangle BNP$ and $\triangle RQD$, however I am having trouble proving those ratios to be equal to 1.