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We first note that the given quadrilateral is a trapezoid, because $60^{\circ }+120^{\circ }=180^{\circ },$ and so the top and bottom sides are parallel. We need to determine the total area of the trapezoid and then what fraction of that area is closest to the longest side.

DETERMINATION OF REGION CLOSEST TO $AD$

Next, we need to determine what region of the trapezoid is closest to side $AD.$ To be closest to side $AD,$ a point inside the trapezoid must be closer to $AD$ than to each of $BC,$ $AB,$ and $DC.$ For a point in the trapezoid to be closer to $AD$ than to $BC,$ it must be below the "half-way mark", which is the midsegment $MN.$ Thus, such a point must be below the parallel line that is $$\frac{1}{2}(50\sqrt{3})=25\sqrt{3}\text{ m}$$above $AD.$

For a point in the trapezoid to be closer to $AD$ than to $AB,$ it must be below the angle bisector of $\angle BAD.$ Similarly, for a point in the trapezoid to be closer to $AD$ than to $DC,$ it must be below the angle bisector of $\angle CDA.$ Define points $X$ and $Y$ to be the points of intersection between the angle bisectors of $\angle BAD$ and $\angle CDA,$ respectively, with the midsegment $MN.$

Solution 1: The slick way:

Connecting $B$ and $C$ to the midpoint of $\overline{AD}$ forms three equilateral triangles as shown below:



$X$ is the midpoint of $\overline{BM}$ and $Y$ is the midpoint of $\overline{CM}.$ Therefore, the region of points closest to $\overline{AD}$ consists of half of triangle $ABM,$ $1/4$ of triangle $BCM$ (since $X$ and $Y$ are midpoints of sides $\overline{BM}$ and $\overline{CM},$ the area of $MXY$ is $1/4$ the area of $BCM$), and half of triangle $CDM$. Each equilateral triangle is $1/3$ of the entire trapezoid, so the region that is closest to $\overline{AD}$ is $$\frac{1}{3}\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{4}\right)=\boxed{\frac{5}{12}}$$of the entire trapezoid. (Solution from user brokenfixer.)

Solution 2: The long way.

AREA OF TRAPEZOID

Label the trapezoid as $ABCD$ and drop perpendiculars from $B$ and $C$ to $P$ and $Q$ on $AD.$ Since $△ABP$ is right-angled at $P$ and $\angle BAP=60^{\circ },$ then $$AP=\frac{1}{2}\cdot 100=50\text{ m}\text{and}BP=\frac{\sqrt{3}}{2}\cdot 100=50\sqrt{3}\text{ m}.$$(We used the ratios in a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle to do these calculations.) By symmetry, $QD=50\text{ m}$ as well.

Also, since $BC$ is parallel to $PQ,$ and $BP$ and $CQ$ are perpendicular to $PQ,$ then $BPQC$ is a rectangle, so $PQ=BC=100\text{ m}.$ Thus, the area of trapezoid $ABCD$ is $$\frac{1}{2}(BC+AD)(BP)=\frac{1}{2}(100+(50+100+50))(50\sqrt{3})$$or $7500\sqrt{3}$ square meters.

AREA OF TRAPEZOID $AXYD$

Lastly, we need to determine the area of trapezoid $AXYD.$ Note that $$\angle XAD=\angle YDA=\frac{1}{2}(60^{\circ })=30^{\circ }.$$Drop perpendiculars from $X$ and $Y$ to $G$ and $H,$ respectively, on $AD.$ We know that $AD=200\text{ m}$ and $XG=YH=25\sqrt{3}\text{ m}.$

Since each of $△AXG$ and $△DYH$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, $$AG=DH=\sqrt{3}XG=\sqrt{3}(25\sqrt{3})=75$$This tells us that the angle bisectors must intersect above $MN,$ since $AG+HD=150$ and $AD=200,$ so $AG+HD<AD.$

Since $XGHY$ is a rectangle (by similar reasoning as for $BPQC$), $$\begin{array}{rl}XY& =GH\\ & =AD-AG-DH\\ & =200-75-75\\ & =50.\end{array}$$Therefore, the area of trapezoid $AXYD$ is $$\frac{1}{2}(AD+XY)(XG)=\frac{1}{2}(50+200)(25\sqrt{3})$$or $3125\sqrt{3}$ square meters.

This tells us that the fraction of the crop that is brought to $AD$ is $$\frac{3125\sqrt{3}}{7500\sqrt{3}}=\frac{25}{60}=\boxed{\frac{5}{12}}.$$