解答:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d31b0ef41bd5ad6e4c05136482cb39dbb7fd3cbb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
证明:(1)如图1,∵△ADE与△ABC都是等边三角形,
∴AC=AB,AE=AD,∠DAE=∠BAC=60°.
∴∠DAE-∠CAD=∠BAC-∠CAD.
即∠CAE=∠BAD.
在△CAE和△BAD中,
∵
,
∴△CAE≌△BAD(SAS).
∴EC=DB(全等三角形的对应边相等);
∴CE+CD=DB+CD=BC=AB,即CE+CD=AB;
(2)CE+CD=AB;
理由如下:如图2,∵△ADE与△ABC都是等边三角形,
∴AC=AB,AE=AD,∠DAE=∠BAC=60°.
∴∠DAE-∠BAE=∠BAC-∠BAE.
即∠CAE=∠BAD.
在△CAE和△BAD中,
∵
,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d31b0ef41bd5ad6e4c05136482cb39dbb7fd3cbb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴△CAE≌△BAD(SAS).
∴EC=DB(全等三角形的对应边相等);
∴CE+AB=DB+BC=CD,即CE+AB=CD.