![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/314e251f95cad1c8703716967c3e6709c83d51f7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
解:(1)∵M是EC的中点,
∴BM=
EC,DM=
EC,(直角三角形斜边上的中线等于斜边的一半),
∴DM=BM.
∵M是EC的中点,
∴MC=
EC,
∴BM=MC=DM,
∴∠1=∠2,∠3=∠4,
∵∠BME=∠1+∠2,∠EMD=∠3+∠4,
∴∠BMD=2(∠1+∠3),
∵△ABC等腰直角三角形,
∴∠BCA=45°,
∴∠BMD=90°,
∴BM=DM且BM⊥DM;
故答案为:BM=DM且BM⊥DM.
(2)成立.
理由如下:延长DM至点F,使MF=MD,连接CF、BF、BD.
在△EMD和△CMF中,
∵
∴△EMD≌△CMF(SAS),
∴ED=CF,∠DEM=∠1.
∵AB=BC,AD=DE,且∠ADE=∠ABC=90°,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d0c8a786c9177f3ebdb816bc73cf3bc79e3d56f7?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴∠2=∠3=45°,∠4=∠5=45°.
∴∠BAD=∠2+∠4+∠6=90°+∠6.
∵∠8=360°-∠5-∠7-∠1,∠7=180°-∠6-∠9,
∴∠8=360°-45°-(180°-∠6-∠9)-(∠3+∠9),
=360°-45°-180°+∠6+∠9-45°-∠9=90°+∠6.
∴∠8=∠BAD.
在△ABD和△CBF中,
∵
,
∴△ABD≌△CBF(SAS),
∴BD=BF,∠ABD=∠CBF.
∴∠DBF=∠ABC=90°.
∵MF=MD,
∴BM=DM且BM⊥DM.