Question

如图23-48,在三角形ACD中,已知角ACD=120度，把三角形ACD绕顶点C逆时针旋转60度得到三角形BCD,画出旋转后的

图形；1连接AB，DE,试判断三角形ABC和三角形CDE的形状；2若AD交EC于N,BE交AC于M,试判断三角形CAN与三角形CBM，线段MN与BD的关系？

Answer 1

1.三角形ABC和三角形CDE都是等边三角形。
因为逆时针旋转后，角ACB与角ECD都是60度，AC=BC，CE=CD
所以两个三角行都是等边三角形。
2.三角形CAN与三角形CBM相似，MN与BD平行

追问

第二小题还不大明白，能不能说仔细一些

追答

不好意思，前面看错了，iggai是这样的：
三角形ACD和三角形BCE是同一个三角形
所以∠CAN=∠CBN
∠ACB=∠ECD=60°
AC=BC
根据ASA可判定两个三角形全等

根据俩个三角形全等，可知CN=CM
因为角ACE=60°
所以三角形CNM为等边三角形
∠MNC=∠ECD=60°
所以MN∥BD

Answer 2

1.两个三角形均为等边三角形
2.∵∠CAN=∠CBM AC=BC
∠ACN=60°=∠BCM
∴△CAN≌△CBM

∴CN=CM
∴△CMN等边
∴MN∥BD (内错角均为60°)

Answer 3

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