在RT△AEC与RT△BFD中∠ADE=∠BDF∴∠DAF=∠DBF∵RT△AEC和RT△CFB中∠CFE=45°-∠DAR∠BCF=90°-(45°+∠DFB)=45°-∠DFB=45°-∠DAE∴∠CAE=∠BCF又∵AC=BC CE=BF∵RT△CHD∽RT△BFD∴∠HCD=∠BF CE=BF∴RT△GEC≌RT△DFB∴BD=CG
为了拿分乱说，无语阿
