解答:(1)解:故答案为:等腰三角形三线合一(或等腰三角形顶角的平分线、底边上的中线、底边上的高互相重合),角平分线上的点到角的两边距离相等.
(2)证明:∵CA=CB,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9345d688d43f8794354041afd11b0ef41ad53af7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴∠A=∠B,
∵O是AB的中点,
∴OA=OB.
∵DF⊥AC,DE⊥BC,
∴∠AMO=∠BNO=90°,
∵在△OMA和△ONB中
,
∴△OMA≌△ONB(AAS),
∴OM=ON.
(3)解:OM=ON,OM⊥ON.理由如下:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/38dbb6fd5266d01697a697a9942bd40734fa35f7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
连接OC,
∵∠ACB=∠DNB,∠B=∠B,
∴△BCA∽△BND,
∴
=
,
∵AC=BC,
∴DN=NB.
∵∠ACB=90°,
∴∠NCM=90°=∠DNC,
∴MC∥DN,
又∵DF⊥AC,
∴∠DMC=90°,
即∠DMC=∠MCN=∠DNC=90°,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/38dbb6fd5266d01697a697a9942bd40734fa35f7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴四边形DMCN是矩形,
∴DN=MC,
∵∠B=45°,∠DNB=90°,
∴∠3=∠B=45°,
∴DN=NB,
∴MC=NB,
∵∠ACB=90°,O为AB中点,AC=BC,
∴∠1=∠2=45°=∠B,OC=OB(斜边中线等于斜边一半),
在△MOC和△NOB中
,
∴△MOC≌△NOB(SAS),
∴OM=ON,∠MOC=∠NOB,
∴∠MOC-∠CON=∠NOB-∠CON,
即∠MON=∠BOC=90°,
∴OM⊥ON.