Question

高分悬赏！！！初一数学题！！！

如图，AB‖CD，已知∠BAF=4∠BAE，∠DCF=4∠DCE，试探讨∠AEC与∠AFC的数量关系，并给出证明。

Answer 1

由条件可得：∠AEC=∠BAE+∠DCE
∠AFC=360°－∠BAF－∠DCF
=360°－4(∠BAE+∠DCE)
=360°－4∠AEC
所以：∠AFC+4∠AEC=360°

Answer 2

∠AFC=360度-4∠AEC

Answer 3

证明：∠AFC=360°—4∠AEC
∵∠BAE+∠DCE=∠AEC
∠FAE=3∠BAE同理∠FCE=3∠ECD
∴∠FAE+∠FCE=3（∠BAE+∠DCE）=3∠AEC
∵四边形AFCE的内角和为360°
∴∠AFC+∠FAE+∠FCE+∠AEC=∠AFC+4∠AEC=360°
∴∠AFC=360°—4∠AEC
应该对吧，好久没有做这样的题了，希望给个好评价！！！O(∩_∩)O~

Answer 4

∠AEC=180-∠CAE-∠ACE
=180-(90-∠BAE)-(90-∠DCE)
=∠BAE+∠DCE
又∠AEC+∠FAE+∠FCE+∠AFC=360
∠AEC+（∠BAF-∠BAE）+（∠DCF-∠DCE）+∠AFC=360
∠AEC+3∠BAE+3∠DCE+∠AFC=360
∠AEC+3(∠BAE+∠DCE)+∠AFC=360
4∠AEC+∠AFC=360

Answer 5

解：过E点做的ab平行线en(n在e左侧)，因为ab//cd,所以en//ab//cd.。所以∠bae=∠aen,因为
∠baf=4∠bae,所以∠fae=3∠bae=3∠aen,同理∠fce=3∠ecd=3∠cen.又因为
∠afc+∠aec+∠fae+∠fce=360'∠
∠afc+∠aec+3∠aen+3∠cen=∠afc+∠aec+3∠aec=∠afc+4∠aec=360
故他们的数量关系为=∠afc+4∠aec=360