1、若∠1=∠2,∠3=∠4,则BD=CE+DE
证明:∵DE‖BC,∴∠2=∠DFB.∵∠1=∠2,∴∠1=∠DFB,∴DB=DF
同理EC=EF.∴DB=DF=DE+EF=DE+CE.
2、题意不清.
3、BD=2CE
证明:延长BA,CE交于点F,作FH⊥BC于H,则FH过点D,(D为ΔFBC垂心)
易证ΔBHD≌ΔFHC,及CE=1/2CF
所以BD=CF=2CE.
4、由①②证③.
过点D作DF‖AC,交BC于F,则DF=BD=CE,易证ΔDFG≌ΔECG,所以DG=GE,G是DE中点.