Solutions; (1) be = (BC-ad)/2 = (14-6)/2 = 4, tan∠B=AE/BE=4 root number 3/4= root number 3, so ∠ B = 60.

(2) Make the intersection point E the vertical line of CD, and intersect with CD at point G.. AB=CD=8, EG=EC*sinC=5, root number 3. S = 1/2××× 5 radical number 3

(3) The trapezoidal area is expressed by SS. SS=AE×(AD+BC)× 1/2，S= 1/4SS，X=4。 DF=CD-4=4。 If the perpendicular of AE intersects with F, intersects with AE at H, and FH is parallel to AD and CE, then H is the midpoint of AE. FH= 1/2×(AD+CE)=8 .AH=EH, FH=FH, ∠AHF=, ∠EHF, so ⊿AHF and ⊿EHF are congruent. So AF=EF