證明：在?$BC$?上截取?$BF=AB，$?連接?$EF$?
∵?$BE$?平分?$∠ABC，$?∴?$∠ABE=∠F BE$?
在?$?ABE$?和?$?F BE$?中
?$ \begin {cases}{AB=BF}\\{∠ABE=∠F BE}\\{BE=BE}\end {cases}$?
∴?$?ABE≌?F BE(S AS)$?
∴?$∠A=∠BFE$?
∵?$AB//CD，$?∴?$∠A+∠D=180°$?
又∵?$∠BFE+∠CFE=180°，$?∴?$∠CFE=∠D$?
∵?$CE$?平分?$∠BCD，$?∴?$∠DCE=∠F CE$?
在?$?CDE$?和?$?CFE$?中
?$ \begin {cases}{∠D=∠CFE}\\{∠DCE=∠F CE}\\{CE=CE}\end {cases}$?
∴?$?CDE≌?CFE(\mathrm {AAS})$?
∴?$CD=CF$?
∴?$BC=BF+CF=AB+CD$?
