在直角梯形ABCD中,AD‖BC,∠ABC=90º,AB=BC,E为AB边上一点,∠BCE=15º,且AE=AD.连接DE交对角线AC于H,连接BH.下列结论:①△ACD≌△ACE;②△CDE为等边三角形;③EH/BE=2;S△EDC/S△EHC=AH/CH.其中结论正确
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/01 18:32:04
![在直角梯形ABCD中,AD‖BC,∠ABC=90º,AB=BC,E为AB边上一点,∠BCE=15º,且AE=AD.连接DE交对角线AC于H,连接BH.下列结论:①△ACD≌△ACE;②△CDE为等边三角形;③EH/BE=2;S△EDC/S△EHC=AH/CH.其中结论正确](/uploads/image/z/2394852-60-2.jpg?t=%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAD%E2%80%96BC%2C%E2%88%A0ABC%3D90%26ordm%3B%2CAB%3DBC%2CE%E4%B8%BAAB%E8%BE%B9%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E2%88%A0BCE%3D15%26ordm%3B%2C%E4%B8%94AE%3DAD.%E8%BF%9E%E6%8E%A5DE%E4%BA%A4%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E4%BA%8EH%2C%E8%BF%9E%E6%8E%A5BH.%E4%B8%8B%E5%88%97%E7%BB%93%E8%AE%BA%EF%BC%9A%E2%91%A0%E2%96%B3ACD%E2%89%8C%E2%96%B3ACE%3B%E2%91%A1%E2%96%B3CDE%E4%B8%BA%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%EF%BC%9B%E2%91%A2EH%2FBE%3D2%EF%BC%9BS%E2%96%B3EDC%2FS%E2%96%B3EHC%3DAH%2FCH.%E5%85%B6%E4%B8%AD%E7%BB%93%E8%AE%BA%E6%AD%A3%E7%A1%AE)
xX[oG++$xޒEHLVUތ6SɤlИ@H $qL^y_虝]NlJPTsfΜ7ggd8sֿUh?\.UM7jsw4=V/A"3/gb@/jTjT5*6tbhب̩HQx쪁G~yjU_z:kȐfi??^ݫ-ԋK;/TݨfxZ!X ^T/.#%"Cդu3O͵a Zܼ[ޫ婽Z!2j3! bfqL!SBjDLE$ݱT/iG`Ԙ =O9?X٬rboA?Ae/8ʉDN#x6n[#Oqg|%=Z铼!1INO29?k^Rq#1Ȁ0:˳6vE˕K<@3BgVo CzD-'A|>$W.Ey,_k@l` Q