Let the area of the quadrilateral ABCD be one, divide the edge ad into three equal parts, the dividing point is E.F, let AE = EF = FD, and divide BC into three equal parts, the dividing point is H.G, so that BH = Hg = GC Connecting EF and GH to prove s quadrilateral efgh = 1 / 3

Let the area of the quadrilateral ABCD be one, divide the edge ad into three equal parts, the dividing point is E.F, let AE = EF = FD, and divide BC into three equal parts, the dividing point is H.G, so that BH = Hg = GC Connecting EF and GH to prove s quadrilateral efgh = 1 / 3

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Proof: connect be, BD, DG, eg
① The first step is to prove that s △ Abe + s △ CDG = 1 / 3
In △ Abe and △ abd, because e is the trisection of AD, AE = ad / 3 and the height h of the two triangles on the side of ad is the same,
S △ Abe = 1 / 2 × AE × H = 1 / 2 × AD / 3 × H = 1 / 3 × (1 / 2 × ad × h) = s △ abd / 3
Similarly, in △ CDG and △ BCD, since G is the trisection point of BC, CG = BC / 3 and the height of two triangles on the side of BC is the same. According to the triangle area formula, s △ CDG = s △ BCD / 3 is obtained
∵ s △ abd + s △ BCD = s quadrilateral ABCD = 1
∴SS△ABE+S△CDG=S△ABD/3+S△BCD/3=1/3（S△ABD+S△BCD）=1/3
② It is proved that s △ beh + s △ DFG = s quadrilateral efgh
∵ in △ beh and △ EGH, h and G are the three equal points of BC, so BH = Hg and the heights of BH and Hg are the same. According to the triangle area formula, s △ beh = s △ EGH
Similarly, in △ DFG and △ EFG, F and E are the triad points of AD, DF = EF and the heights of DF and EF are the same. According to the triangle area formula, s △ DFG = s △ EFG
S △ beh + s △ DFG = s △ EGH + s △ EFG = s quadrilateral efgh
According to the conclusion of ①, it is concluded that s quadrilateral BedG = s quadrilateral ABCD - ① = 2 / 3
According to the conclusion of ②, it is concluded that efgh = BedG / 2 = 1 / 3
It's over