|
|
Last edited by hbghlyj 2020-1-31 11:35
在$\triangle ABC$中,$\angle A=\angle ABC=\angle ADB=70°,CD=BE,$求证:$\angle BDE=50°$
①取$\triangle BCD$外心F,$\angle CFD=2\angle CBD=60°$,$\triangle CFD$为正三角形,$\angle DFB=2\angle ACB=80°,\angle EBF=\frac12(180°-(50°-30°))=80°,\therefore\angle DFB=\angle EFB$,DEF共线,$\angle BDE=180°-70°-50°=60°$
②见图 |
|
