Question

初二几何思考题,如图,在△ABC中,∠ABC=45度,CD⊥AB于D,BE平分∠ABC,且BE⊥AC于E,与CD交于点F

Answer 1

证明:(1)CD垂直AB,BE垂直AC,则∠ACD=∠FBD(均为∠A的余角).

又∠ABC=45°,则:CD=BD;∠CDA=∠BDF=90°.

∴ ⊿CDA≌⊿BDF,BF=AC.

(2)∠ABE=∠CBE;BE=BE;∠BEA=∠BEC=90度.

则⊿BEA≌⊿BEC(ASA),AE=CE,AC=2CE,故BF=2CE.

(3)BD=CD,BH=CH,则DH垂直BC,即DH垂直平分BC,CG=BG.

故∠GCB=∠GBC=(1/2)∠ABC=22.5°,∠EGC=∠GCB+∠GBC=45°,得:EG=CE=AE.

EG>GF,所以AE>GF.