Question

△ABC中,∠ABC=45°,CD⊥AB于D,BE平分∠ABC,且BE⊥AC于E,与CD相交于点F,

Answer 1

⑴∵∠ABC=45°,CD⊥AB,∴∠DCB=45°,∴DB=DC

∵∠ABE+∠A=∠ACD+∠A=90°,

∴∠ABE=∠ACD

∴RT△DBF≌RT△DCA

∴BF=AC

⑵∵BE平分∠ABC,BE⊥AC,AE为公共边,

∴RT△BEC≌RT△BEA

∴CE=AE=1/2AC,

∴CE=1/2BF.

⑶△DBC是等腰直角三角形,H是斜边中点,∴DM⊥BC,BM=DM=√2/2CD,

∵∠GBM=∠GBD=∠ACD,

∴RT△BMG∽RT△CDA,

∴BG∶AC=BM∶CD=√2/2,又CE=1/2AC

∴BG=√2CE.