证:延长BE交AC于F
AE=AE,角BAE=角FAE,角AEB=角AEF=90,则三角形AEB全等于AEF,角AFB=角ABF,BE=EF=BF/2
又,角AFB=角C+角FBC,角ABF+角FBC=3角C,所以角ABF=角AFB=2角C,从而角FBC=角C,BF=FC=AC-AF=AC-AB
所以BE=BF/2=(AC-AB)/2
证:延长BE交AC于F
AE=AE,角BAE=角FAE,角AEB=角AEF=90,则三角形AEB全等于AEF,角AFB=角ABF,BE=EF=BF/2
又,角AFB=角C+角FBC,角ABF+角FBC=3角C,所以角ABF=角AFB=2角C,从而角FBC=角C,BF=FC=AC-AF=AC-AB
所以BE=BF/2=(AC-AB)/2