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20 14 detailed explanation of 25 problems in the second model mathematics of grade three in Baoshan District and Jiading District.
Let AH⊥BC be in H,FG⊥BC be in G, in △ABC, AB=AC= 10, cosB=4/5.

∴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.