(2011•沈阳)已知,△ABC为等边三角形,点D为直线BC上一动点(点D不与B、C重合).以AD为边作菱形ADEF,使∠DAF=60°,连接CF.(1)如图1,当点D在边BC上时,①求证:∠ADB=∠AFC;②请直接判断结
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 22:32:45
![(2011•沈阳)已知,△ABC为等边三角形,点D为直线BC上一动点(点D不与B、C重合).以AD为边作菱形ADEF,使∠DAF=60°,连接CF.(1)如图1,当点D在边BC上时,①求证:∠ADB=∠AFC;②请直接判断结](/uploads/image/z/10411993-1-3.jpg?t=%EF%BC%882011%26%238226%3B%E6%B2%88%E9%98%B3%EF%BC%89%E5%B7%B2%E7%9F%A5%2C%E2%96%B3ABC%E4%B8%BA%E7%AD%89%E8%BE%B9%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E7%82%B9D%E4%B8%BA%E7%9B%B4%E7%BA%BFBC%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9%EF%BC%88%E7%82%B9D%E4%B8%8D%E4%B8%8EB%E3%80%81C%E9%87%8D%E5%90%88%EF%BC%89%EF%BC%8E%E4%BB%A5AD%E4%B8%BA%E8%BE%B9%E4%BD%9C%E8%8F%B1%E5%BD%A2ADEF%2C%E4%BD%BF%E2%88%A0DAF%3D60%C2%B0%2C%E8%BF%9E%E6%8E%A5CF%EF%BC%8E%EF%BC%881%EF%BC%89%E5%A6%82%E5%9B%BE1%2C%E5%BD%93%E7%82%B9D%E5%9C%A8%E8%BE%B9BC%E4%B8%8A%E6%97%B6%2C%E2%91%A0%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0ADB%3D%E2%88%A0AFC%EF%BC%9B%E2%91%A1%E8%AF%B7%E7%9B%B4%E6%8E%A5%E5%88%A4%E6%96%AD%E7%BB%93)
xSaoP1&F,&0GC0ю W`Ke+m?}Կ}%O}ys96OۈFym=fcoMG1'DJԴz)5U->^ŵv-W Y!#uԷ kTݗD`%IBgB餥LbkSV,A¨4(6
~h#l͚Yte HTB|2rh7FJ5tns/#:Auk qrACfL
Z]dJ}&OP'쾃
9.7@oa{̎/r
J> Le1ZU-&6&65!n0A#DH9.(j&4FbkP*8[;'o{/z?KdIY"J%kxvǀa:LԸfÐOԬ]wQv#䬰%[6=G^/bQw<
(2011•沈阳)已知,△ABC为等边三角形,点D为直线BC上一动点(点D不与B、C重合).以AD为边作菱形ADEF,使∠DAF=60°,连接CF.(1)如图1,当点D在边BC上时,①求证:∠ADB=∠AFC;②请直接判断结
(2011•沈阳)已知,△ABC为等边三角形,点D为直线BC上一动点(点D不与B、C重合).以AD为边作菱形ADEF,使∠DAF=60°,连接CF.
(1)如图1,当点D在边BC上时,
①求证:∠ADB=∠AFC;②请直接判断结论∠AFC=∠ACB+∠DAC是否成立;
(2)如图2,当点D在边BC的延长线上时,其他条件不变,结论∠AFC=∠ACB+∠DAC是否成立?请写出∠AFC、∠ACB、∠DAC之间存在的数量关系,并写出证明过程;
(3)如图3,当点D在边CB的延长线上时,且点A、F分别在直线BC的异侧,其他条件不变,请补全图形,并直接写出∠AFC、∠ACB、∠DAC之间存在的等量关系.
(2011•沈阳)已知,△ABC为等边三角形,点D为直线BC上一动点(点D不与B、C重合).以AD为边作菱形ADEF,使∠DAF=60°,连接CF.(1)如图1,当点D在边BC上时,①求证:∠ADB=∠AFC;②请直接判断结
(1)ab=ac,ad=af,角bad=caf,所以,三角形bad与caf全等,因此得证;相等.
剩下的没时间回答,手机写太麻烦