Question

如图所示,点E,D分别在△ABC的边AB,BC上,CE和AD交于点F,若S△ABC=1,S△BDE=S△DCE=S△ACE，则S△AED等于

详细解答过程~在线等10分钟

Answer 1

因为S△BDE=S△DCE，所以D为BC中点；设D到AB边距离为h，则C到AB边距离为2h；
有1/2×AE×2h=1/2×h×BE，故BE=2AE；S△AED=1/2×h×AE，S△ABC=1/2×2h×AB=1；
AB=3AE；故S△=1/6S△ABC=1/6

Answer 2

S△BDE=S△DCE推得BD=DC
S△BDE=S△DCE=S△ACE推得S△ACE=S△BEC/2推得AE=AB/3
所以S△AED=S△ABD/3=(S△ABC/2)/3=S△ABC/6=1/6

Answer 3

S△BDE = S△DCE推得BD = DC
S△BDE = S△大商所= S△ACE按S△ACE = S△BEC / 2推AE = AB / 3
△AED = S△ABD / 3 =（S△ABC / 2）/ 3 = S△ABC / 6 = 1/6

Answer 4

AED=ACDE-ADC