37,53 and trapezoid

Extend $AD$ and $BC$ and suppose they meet at $E$ . $E=90°$ as $D=37°$ and $C=53°$ . $EN=1004$ as it is the median of triangle $DEC$ . $EM = 500$ (same reason). Hence, $MN=504$ .

$\tan 37=\dfrac{h}{x}$ $\implies$ $h=x\tan 37$ $\color{#D61F06}(1)$

$\tan53=\dfrac{h}{1008-x}$ $\implies$ $h=(1008-x)(\tan 53)$ $\color{#D61F06}(2)$

$h=h$ Equate $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ .

$x\tan 37= \tan 53 (1008-x)$

$x\left(\dfrac{\tan 37}{\tan 53}\right)=1008-x$

$0.5678x=1008-x$

$1.5678x=1008$

$x\approx 642.94$

It follows that $h=(642.94)(\tan 37) \approx 484.49$

$y=1004-(1008-x)-500=1004-(1008-484.49)-500=138.94$

By pythagorean theorem, we have

$MN=\sqrt{(484.49)^2+(138.94)^2}=$ $\boxed{504}$

$\text{Draw perpendiculars from A, M, B, on DC with feet at P, Q, R .}\\ The ~ height ~ h =\dfrac{2008-1000}{Cot37+Cot53}=484.47.\\ CP=h*Cot53=365.07. \\ \therefore ~ PN=\dfrac {CD} 2-CP, ~~ and ~~AM=\dfrac{AB} 2.\\ \therefore ~QN=PN - AM=138.93. ~~~\implies ~ MN=\sqrt{h^2+QN^2}=503.9967518.\\ \text{SINCE answer box need answer in integer, } MN=\Large ~~~\color{#D61F06}{504}$

$~~~~~~OR ~~~~~~$

$\text{ Let BEC=53 degrees. So ABEC is an isosceles trapezoid. M, L are midpoints of AB and CE.}\\ From ~ ABDC, ~~ the ~ height ~ h =\dfrac{2008-1000}{Cot53+Cot37}=484.47.\\ ED=h*(Cot37 - Cot53)\\ \text{Line CE, E is pulled through ED, its midpoint is pulled by } \dfrac {ED} 2.\\ \dfrac {ED} 2=LN=138.9195 . ~~~So ~ MN=\sqrt {h^2+LN^2}=503.9967518.$