It is known that e is the edge of the parallelogram ABCD, AB extends to the upper point, de intersects BC with F. proof: the area of the triangle ABF is equal to the area of the triangle efc Mathematics in grade two

It is known that e is the edge of the parallelogram ABCD, AB extends to the upper point, de intersects BC with F. proof: the area of the triangle ABF is equal to the area of the triangle efc Mathematics in grade two

Make vertical lines en and am respectively through E and a, and cross BF to N and m
According to DC / / EB, FC: BF = FD: Fe
According to AD / / BF, EF: DF = EB: ab
From AM / / BN, AB: EB = am: en
So FC: BF = am: en
The area of triangle ABF is equal to 1 / 2am * BF
The area of triangle EFC is equal to 1 / 2EN * CF
So the area of triangle ABF is equal to the area of triangle efc

As shown in the figure, e is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC at point F,

∵AD∥CF
S ⊿ AFC = s ⊿ DCF
∵ s ⊿ AFC / s ⊿ ABF = CF / BF
∵ s ⊿ DCF / s ⊿ EFC = DF / ef
∵AE∥CD
∴CF／BF=DF／EF
∴S⊿AFC／S⊿ABF=S⊿DCF／S⊿EFC
∴S△ABF=S△EFC

E is a point of the extension line of AB of parallelogram ABCD, and de intersects BC with F. if the area of triangle ABF = 3, the area of triangle EFC is calculated

ABCD is a parallelogram
S △ ADF = 1 / 2S parallelogram ABCD
S △ ABF + s △ CDF = 1 / 2S parallelogram ABCD
∵ s △ ECD = 1 / 2S parallelogram ABCD
∴S△ABF+S△CDF=S △CEF +S△CDF
∴S△CEF =S △ABF
∵S△ABF =3
∴S△CEF=3