Rt△ABC是边长比为3:4:5的直角三角形,所以按此比例计算即可
AB = 10
AD = 8 - 0.5 = 7.5 = 5份
DE = 3份 = 4.5
AE = 4份 = 6
AA1 = 2AE = 12
BA1 = 12 - 10 = 2
点A1、F、D作为截点,由梅涅劳斯定理（不用定理作BG//AC交A1D于G）
BA1/A1A * AD/DC * CF/FB = 1
即 2/12 * 7.5/0.5 * CF/FB = 1
CF/FB = 2/5
CF = BC*2/7 = 12/7
四边形DEBF的面积 = S△ABC - S△CDF - S△AED
= 6*8/2 - 0.5*12/7*1/2 - 6*4.5/2
= 24 - 3/7 - 27/2
= 141/14