1.已知△ABC,AD平分∠BAC,CE⊥AD交AB于E,EF‖BC,交AC于F.求证∠FEC=∠CED

2个回答

  • 1.设AD跟CE的交点为O

    ∵AD平分∠BAC,∴∠EAO=∠CAO

    ∵CE⊥AD,∴∠AOE=∠AOC=90度

    ∴∠AEO=∠ACO,∴△AOE≌△AOC,∴OE=OC

    ∵CE⊥AD,∴∠EOD=∠COD

    ∵OD=OD,∴△EOD≌△COD,∴∠OED=∠OCD

    ∵EF‖BC,∴∠FEC=∠ECD

    ∴∠FEC=∠CED

    2.∵ED垂直平分AC,∴CD=AC,∠CDE=∠ADE=90度,ED=ED,

    ∴△CDE≌△ADE,∴∠EAD=∠ECD

    ∵∠B=90度,∠EAB:∠BAC=2:5,∠EAD=∠ECD

    ∴∠EAD+∠ECD+∠EAB=90度,∠EAB:∠BAC:∠ECD=2:5:3

    ∴∠C=33度45分

    3.设AD跟EF的交点为O

    ∵AD是EF的垂直平分线,∴EO=FE,∠EOD=∠FOD=90度

    ∵OD=OD,∴△EOD≌△FOD,∴∠OED=∠OFD,OE=OF

    ∵AO=AO,∠AOE=∠AOF=90度,∴△AOE≌△AOF

    ∴∠AEO=∠AFO,∴∠AED=∠AFD

    ∴∠BED=∠DFC

    绝对正确~给分吧~不给不厚道~