It's also nice and easy problem.We must prove that $ TP^2=TR \cdot TS=TB \cdot TC $ , where $ T=MN \cap BC $ ( w.l.o.g $ B $ lies between $ T $ and $ C $ ) .Let $ MN \cap AP=F $ , since $ AMPN $ is p.g.m so $ F $ is common midpoint of $ MN $ and $ AP $ and $ AN=PM , AM=BN $ . From Meneleus theorem we have (1) In the triangle $ APC $ the line $ TFN $ is transerval ,so $ \frac{TP}{TC} \cdot \frac{CN}{NA} \cdot \frac{AF}{FM} =1 \implies \frac{TP}{TC}=\frac{AN}{CN} $ . (2)In the triangle $ ABC $ the line $ TMN $ is transerval, so $ \frac{TB}{TC} \cdot \frac{CN}{NA} \cdot \frac{AM}{BM} \implies \frac{TP}{TB} = \frac{AM}{MB} $ ( I use that (1) ).Hence we have to prove that $ \frac{AM}{BM}=\frac{CN}{AN} $ or $ \frac{PN}{MB}=\frac{CN}{PM} $ , which is true ,because the triangle $ BMP $ and $ CPN $ are similar.DONE!

Very nice problem! Here is another attempt. $ T=MN \cap BC $ ( w.l.o.g $ B $ lies between $ T $ and $ C $ ) We have $\frac{TM}{TN} = \frac{TP}{TC}$ (since TMP and TNC are similar) and $\frac{TM}{TN} = \frac{TB}{TP}$ (since TMB and TNP are similar) So, $\frac{TB}{TP} = \frac{TP}{TC}$ (since TMP and TNC are similar) or, $TP^2 =TB*TC$ From $TB*TC=TR*TS$ Then, $TP^2=TR*TS$ And the result follows.

I have the same solution as --Fermat--,an easy problem

The configuration is boring Also, the condition that $MN, BC$ are not parallel is a dead giveaway. It's totally unnecessary and the problem is valid without it, so why would they put it there? Obviously we're supposed to intersect them. Then angle chasing turns out to be really really annoying, and Menelaus is the next choice because of how the diagram looks, and it works.