Question

如图AB∥CD,BE,CE分别是∠ABC,∠BCD的平分线点E在AB上,求证BC=AB+CD

Answer 1

E在AD上.

延长BE交CD的延长线于F,

∵AB∥CD,∴∠ABE=∠F,

∵∠ABE=∠EBC,∴∠F=∠EBC,∴BC=FC.

∵AB∥CD,∴∠ABC+∠BCD=180°,

∵BE、CE平分∠ABC、∠BCD,∴∠EBC+∠ECB=1/2(∠ABC+∠BCD)=90°,

∴CE⊥BF,∴BE=EF(等腰三角形三线合一)

∵∠A=∠EDF,∠ABE=∠F,

∴ΔABE≌ΔDFE,

∴AB=DF,

∴BC=CF=DF+CD=AB+CD.