如图,△ABC中,∠C=90°,AC=BC,AD平分∠CAB交BC于点D,DE⊥AB,垂足为E,且AB=6cm,则△DE
1个回答

∵AD平分∠CAB交BC于点D

∴∠CAD=∠EAD

∵DE⊥AB

∴∠AED=∠C=90

∵AD=AD

∴△ACD≌△AED.(AAS)

∴AC=AE,CD=DE

∵∠C=90°,AC=BC

∴∠B=45°

∴DE=BE

∵AC=BC,AB=6cm,

∴2BC 2=AB 2,即BC=

AB 2

2 =

6 2

2 =3

2 ,

∴BE=AB-AE=AB-AC=6-3

2 ,

∴BC+BE=3

2 +6-3

2 =6cm,

∵△DEB的周长=DE+DB+BE=BC+BE=6(cm).

另法:证明三角形全等后,

∴AC=AE,CD=DE.

∵AC=BC,

∴BC=AE.

∴△DEB的周长=DB+DE+EB=DB+CD+EB=CB+BE=AE+BE=6cm.

故选B.