I'm not sure why no one has posted a solution here yet. The problem is not very hard. Let the lines through $P,Q$ parallel to $\overline {BC}$ and $\overline{AD}$ intersect $\overline {CD}$ in $P',Q'$ and let the parallel line through $N$ intersect $\overline{AB},\overline{CD}$ in $Q,L$. Assume WLOG that $CD=4$, and let $r=BC$,$s=AD$, $y=CL$. We need to show that $LN \le LQ$. Because of the similarity of triangles $CQQ'$ and $CNM$, we have that $LN=ry/2$. Similarly, $LN=(4-y)s/2$. Solving yields $y=4s/(r+s)$ and $LN=2rs/(r+s)$. Now $LQ=(sy+r(4-y))/4 =(r^2+s^2)/(r+s)$, so the desired result follows from the AM-GM inequality.

Let's denote $S_{F}$ the area of figure $F$. It is obvious one of the angles $BCD$ or $CDA$ is acute. Therefore we easily obtain that point $N$ is between lines $CD$, $BC$, $AD$. Suppose point $N$ is outside or the quadrilateral $ABCD$ $\Rightarrow$ $S_{TBC}+S_{TCD}+S_{TDA}>S_{ABCD}$. $BQ=QM\Rightarrow S_{BNQ}=S_{NQM}$, $S_{BQC}=S_{QCM}$ $\Rightarrow S_{NBC}=S_{NCM}$ Similarly $S_{NDM}=S_{NDA}$. $CM=MD \Rightarrow S_{NBC}=S_{NCM}=S_{NMC}=S_{NDA}=X$ $S_{TBC}+S_{TCD}+S_{TDA}=4X$ Let the distance between $N$ and $AD$ be $y$ and between lines $BC$ and $AD$ be $h$. $S_{AND}=X=\frac{1}{2}yAD=\frac{1}{2}(h-y)BC \Rightarrow y=\frac{h BC}{BC+AD}$ $S_{ABCD}<S_{TBC}+S_{TCD}+S_{TDA}=4X=4\frac{1}{2}\frac{hBC}{BC+AD}AD=\frac{2h AD BC}{AD+BC}<h\frac{AD+BC}{2}=S_{ABCD}$ we used here AM-GM. Contradiction.

polya78 wrote: I'm not sure why no one has posted a solution here yet. The problem is not very hard. Let the lines through $P,Q$ parallel to $\overline {BC}$ and $\overline{AD}$ intersect $\overline {CD}$ in $P',Q'$ and let the parallel line through $N$ intersect $\overline{AB},\overline{CD}$ in $Q,L$. Assume WLOG that $CD=4$, and let $r=BC$,$s=AD$, $y=CL$. We need to show that $LN \le LQ$. Because of the similarity of triangles $CQQ'$ and $CNM$, we have that $LN=ry/2$. Similarly, $LN=(4-y)s/2$. Solving yields $y=4s/(r+s)$ and $LN=2rs/(r+s)$. Now $LQ=(sy+r(4-y))/4 =(r^2+s^2)/(r+s)$, so the desired result follows from the AM-GM inequality. you said that $\triangle PQQ'\sim \triangle PMN$ means $QQ'||MN\implies MN||AD$ how do you proved that?

we will use the coordinates $(D;\vec{DA},\vec{DC})$ we put D$(0,0)$ A$(1,0)$ C$(0,c)$ B$(b,c)$ and thus M$(0,c/2)$ P$(1/2,c/4)$ Q$(\frac{b}{2},3c/4)$ solve for N N$(\frac{2b}{b+1},\frac{bc}{b+1})$ now it's clear the the coordinates of $N$ are positive so it suffices to show that $N$ is to the left to $AB$ and down of $CB$ the equation of $AB$ is $y=\frac{cx-c}{b-1}$ so the if a point is to the left of that line it must satisfy the inequality: $y<\frac{cx-c}{b-1}$ if $b<1$ or $y>\frac{cx-c}{b-1}$ if $b>1$

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which is clear if we put $N$ into the inequality and because $c>\frac{bc}{b+1}$ $N$ is down of $CB$ and we win

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It's obviously sufficient to show that $N$ lies on the same side of $AB$ as $D$. We use barycentric coordinates with respect to $\triangle ABD$. Let $C=(-c,1,c)$. Take $c>0$, then we have to show that the $z$-coordinate of $N$ is non-negative. Indeed, $M=(-c:1:1+c),P=(2-c:1:1+c),Q=(-c:3:1+c)$, so $\overline{DP}: x=(2-c)y$ and $\overline{CQ}: 0=(1-2c)x+cy-2cz$. Intersecting these two lines yields $N=\left(\frac{c(2-c)}{c+1},\frac{c}{c+1},\frac{(c-1)^2}{c+1}\right)$ and the last coordinate is $\geq 0$, as desired.

A nice problem for coordinates! anyway D=(0,0) A=(1,0) and C=(0,m) and B=(n,m). now M=(0, m/2) and P=(1/2, m/4) and Q=(n/2, 3m/4) so DP is y=mx/2, and CQ slope is m/2n, now if we solve for N, we get N=(2n/(n+1), mn/(n+1)), now we need to show that N is below BC, so we need to prove m>(mn)/(n+1) which follows because 1>n/(n+1) yay