Question

如图,AD平行于BC,AE,BE分别平分角DAB,角ABC,CD过点E,求证:AB=AD+BC

Answer 1

证明：过E作EF//AB交AD于点F,∵EF//AB∴∠BAE=∠AEF∵∠BAE=∠DAE∴∠AEF=∠DAE∴△FAE是等腰三角形∴AF=EF∵AB//CD∴EF//CD∴∠CDE=∠DEF∵∠ADE=∠CDE∴∠DEF=∠ADE∴△FDE是等腰三角形∴DF=EF∴F是AD的中点,且AD=...