∴CH=BH=8,AH=6，

(1)BD=x, DE=3(E is to the right of D),

∴BE=x+3，

EF∑AC，

∴EF/AC=BE/BC，

∴y=EF=AC*BE/BC=5(x+3)/8，

By be

(2) BF = ef, and △ BDF is a right triangle, which is divided into two cases:

1)BD⊥DF,△BDF∽△BHA, one DF=3BD/4=3x/4,

According to Pythagorean Theorem, BF=5x/4=5(x+3)/8, x=3,

∴△ area of ∴△BDF =( 1/2)3*9/4=27/8.

2)bf⊥fd,bf=4x/5=5(x+3)/8,32x=25x+75,7x=75,x=75/7，

DF=3x/5=45/7，

∴△ area of ∴△BDF = (1/2) (75/7) (45/7) = 3375/98.

(3)MN passes through the center of gravity of △DEF, MN∨BC passes through FD and FE at m, n,

∴FM/MD=FN/NE=2,MN=(2/3)DE=2，

When D and B coincide, the positions of M and N are M0 and N0, respectively.

When e and c coincide, the positions of m and n are M 1 and N 1 respectively.

From m0n 0 ∨= m 1n 1 we know that the quadrilateral m0n 1m 1 is a parallelogram.

BM0=5/8，CN 1= 10/3，

∴ parallelogram m0n 1m 1 area = 2 (13-5/8) * 3/5 =13/4.