Question

如图,在平行四边形ABCD中,E为AB延长线上任意一点,DE交BC于F,试判断三角形ABF面积与三角形EFC的面积关系

并证明

Answer 1

解：S△ABF = S△EFC
证明如下：
过F作MN⊥AE于M,交CD于N.作ES⊥BC于S,过D作DT⊥BC于T,
设AD = 1,AB=a,BE=b,DT=h,则MN=h/a.
因为AE//CD
所以△BFE∽△CFD 所以BF:CF = MF:NF = ES:DT = b:a
所以MF = b/(a+b) * (h/a) , ES = bh/a, CF = a/(a+b)
因为S△ABF = (1/2) * AB * FM = (1/2)* a * b/(a+b) * (h/a)= (bh) / [2(a+b)]
S△EFC = (1/2) * CF * ES = (1/2) * a/(a+b) * (bh/a) = (bh) / [2(a+b)]
所以S△ABF = S△EFC