已知△ABC中,∠BAC=2∠B,AB=2AC,AE平分∠BAC.求证∠C=90°

网友回答

证明：过点E作EF⊥AB于D∵AE平分∠BAC∴∠CAE＝∠BAE＝1/2∠BAC∵∠BAC＝2∠B∴∠B＝1/2∠BAC∴∠B＝∠BAE∴AE＝BE∵EF⊥AB∴∠AFE＝∠BFE＝90,AF＝BF （等腰三角形三线合一）∴AB＝2AF∵AB＝2AC∴AC＝AF∵AE＝AE∴...