证明:延长DE交CB延长线于
AD//BC
∠DAE=∠EBF,∠AED=∠BEF,AE=BE
△ADE≌△EBF(ASA)
AD=BF,∠ADE=∠EFB
又DE平分∠ADC
∠ADE=∠EDC,∠ADE=∠EFB
∠EDC=∠EFB
DC=FC
FC=BF+AD,
AD=BF
DC=AD+AD
证明:延长DE交CB延长线于
AD//BC
∠DAE=∠EBF,∠AED=∠BEF,AE=BE
△ADE≌△EBF(ASA)
AD=BF,∠ADE=∠EFB
又DE平分∠ADC
∠ADE=∠EDC,∠ADE=∠EFB
∠EDC=∠EFB
DC=FC
FC=BF+AD,
AD=BF
DC=AD+AD