8.AD∥BC,设AD=a,AD与BC的距离为h,
AD:BC=2:3,
所以BC=3a/2,
梯形ABCD的面积=5ah/4=112,
所以ah=448/5.
延长CH交DA延长线于A1,延长DH交CB延长线于B1.
AE:EF:FB=1∶3:3,
所以A1A/BC=AE/EB=1/6,
A1A=BC/6=a/4,A1D=5a/4.
B1B/AD=BF/FA=3/4,
B1B=3a/4,B1C=9a/4.
所以A1H/HC=A1D/B1C=5/9,
所以S△CDH/S△CDA1=9/14,
S△CDA1=(1/2)*5a/4*h=5ah/8,
所以S△CDH=5ah/8*9/14=45ah/112=(45*448/5)/112=36.