In triangle CAB, PMX is the transversal. Applying Menalaus' Theorem we have $\frac{PA}{AC} = \frac {BM}{MC}$ In triangle CAD, PYN is the transversal. Applying Menalaus' Theorem we have $\frac{PA}{AC} = \frac {AN}{DN}$ Conclusion follows.

Actually, I am pretty sure it should be $\frac{PA}{AC}= \frac{DM}{AM}$ and $\frac{PA}{PC} = \frac{NC}{BM}$. This means that we are done by Thales.

Easy but very nice.Draw $CL \parallel PM$ and $AZ \parallel PN$ with $L,Z$ on $AD,BC$(possibly extended) to conclude that $\frac{DM}{AM}=\frac{CN}{BN}=\frac{PC}{AP}$.So $MN \parallel AB$. Maths is the doctor of science..... Sayantan...

From theorem Menelaus $\frac{Cn} {BN} \frac{BX} {AX} \frac{AP} {PC}=1$ and $\frac{CY} {DY} \frac{DM} {AM} \frac{AP} {PC}=1$ We know $AX=BX$ and $CY=DY$ $\Rightarrow$ $\frac{CN} {BN} \frac{AP} {PC}=1$; $\frac{DM} {AN} \frac{AP} {PC}$ $\Rightarrow$ $\frac{DM} {AM}=\frac{CN} {BN}$ $\Rightarrow$ $\Rightarrow$ $\frac{AM} {DM}=\frac{BN} {CN}$ $\Rightarrow$ $\frac{AM} {DM}+1=\frac{BN} {CN}+1$ $\Rightarrow$ $\frac{AD} {DM}=\frac{BC} {CN}$ $\Rightarrow$ $MN\parallel AB$.

We actually need $\frac{AX}{BX}=\frac{DY}{CY}$ only! Best regards, sunken rock

Goutham wrote: Let $ABCD$ be a trapezium with $AB\parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$; and let $X, Y$ be the midpoints of $AB, CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$. Prove that $MN\parallel AB$. Let $\infty$ be the point at infinity of $\ell_{AB}$.Let $P\infty$ cut $AD,BC$ at $G,H$ respectively\[-1=(A,B,X,\infty)\overset{P}{=}(B,C,N,H)\]\[-1=(D,C,Y,\infty)\overset{P}{=}(D,A,M,G)\].Clearly we have $AB\cap CD\cap GH=\infty$, since ratio is bijective whence $MN$ passes through $\infty \blacksquare $

Purely projective restatement: Quote: Let \(ABCD\) be a quadrilateral. Let \(X\in AB\) and \(Y\in CD\) such that the lines \(AD,BC,XY\) concur at a point \(F\).For any point \(P\) on line \(AC\), let \(N=PX\cap BC\) and \(M=PY\cap AD\). Then the lines \(AB,DC,MN\) concur at a point \(E\).