如图所示,△ABC中,BD⊥AC于点D,AE平分∠BAC,AE交BD于点F,∠ABC=90°.
(1)求证:∠BEF=∠BFE;
(2)若BC=80cm,BE:EC=3:5,AC=100cm,求S
△AEC和S
△ABC.
(1)如图,∵AE平分∠BAC,∴∠1=∠2,∵BD⊥AC,∠ABC=90°,∴∠1+∠BEF=∠2+∠AFD=90°,∴∠BEF=∠AFD,∵∠BFE=∠AFD(对顶角相等),∴∠BEF=∠BFE;(2)∵BC=80cm,BE:EC=3:5,∴EC=80×53+5=50cm,由勾...