1) As shown in the figure, if DT is perpendicular to BC, AB = DT = TC = 8；; Because BC= 14 and BT=BC-TC= 14-8=6, then AD = BT = 6；;

2) Continue to make FQ perpendicular to BC.

Because the angle EPF = 90°, PE=PF, PEF= PFE = 45° (QFC= QFN= QNF= C = 45). In the right triangle NFC, FQ is perpendicular to NC, so FQ=NC/2, so NC = 2fq = 2be = 2.

And BM = be = 8-X.

Because BM+NC=BC+NM, that is, (8-x)+2(8-x)= 14+y and y= 10-3x.

3) When MN is 2, that is, y=2, x=8/3, that is, AE=8/3. EF=ER+RF=AD+DR=6+8/3=26/3。

The area of trapezoidal AEFD = (ad+ef) * AE/2 = (6+26/3) * (8/3)/2 =176/9.