如图,△ABC中,AB>AC,AD平分∠BAC,CD⊥AD,点E是BC的中点. 求证: (1)DE∥AB; (2)DE=1/2(AB-AC).

问题描述:

如图,△ABC中,AB>AC,AD平分∠BAC,CD⊥AD,点E是BC的中点.
求证:

(1)DE∥AB;
(2)DE=

1
2
(AB-AC).

证明:如图,延长CD交AB于点F,∵AD平分∠BAC,∴∠CAD=∠FAD,∵CD⊥AD,∴∠ADC=∠ADF=90°,在△ADC和△ADF中,∠CAD=∠FADAD=AD∠ADC=∠ADF=90°,∴△ADC≌△ADF(ASA),∴CD=DF,AC=AF,∵点E是BC的中点...