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[Kneel down] Math and geometry problems in the second day of junior high school. Give a detailed proof. Sit down and wait.
Let BE=a=GF first, and then EF=GB=2a.

Connect AF. The analysis shows that the area of triangular ACF = quadrilateral ACGF- triangular CFG.

Quadrilateral ACGF= trapezoid ABGF+ triangle ABC

Trapezoidal abgf = (gf+ab) × gb÷ 2 = (a+3 )× 2a ÷ 2 = a× a+3a.

Triangle ABC=AB×BC÷2=3×6÷2=9.

Triangle CFG = GF× CG ÷ 2 = GF× (CB+BG) ÷ 2 = A× (6+2A) ÷ 2 = A× A+3A.

So quadrilateral ACGF=(a×a+3a)+9.

The area of triangle ACF =(a×a+3a)+9-(a×a+3a)=9.

To sum up, the area of triangular ACF =9.

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