如图10所示,已知Rt△ABC≌Rt△ADE,∠ABC=∠ADE

问题描述:

如图10所示,已知Rt△ABC≌Rt△ADE,∠ABC=∠ADE

是不(2)证法一:连接CE,∵Rt△ABC≌Rt△ADE,∴AC=AE.∴∠ACE=∠AEC(等边对等角).又∵Rt△ABC≌Rt△ADE,∴∠ACB=∠AED.∴∠ACE-∠ACB=∠AEC-∠AED.即∠BCE=∠DEC.∴CF=EF.证法二:∵Rt△ABC≌Rt△ADE,∴AC=...