①已知AB=BC=CD,O为DE的中点即DO=EO,∴AE-CO=AB+BC+EO=14-6=8,BC+EO=CD+DO=CO=6,∴AB=AB+BC+EO-(BC+EO)=AE-CO-(BC+EO)=AE-CO-CO=14-6-6;②已知∠DOB=16∠AOB,∠BOE=23∠BOC,∠DOB与∠BOE互余,∴得:(...
①已知AB=BC=CD,O为DE的中点即DO=EO,∴AE-CO=AB+BC+EO=14-6=8,BC+EO=CD+DO=CO=6,∴AB=AB+BC+EO-(BC+EO)=AE-CO-(BC+EO)=AE-CO-CO=14-6-6;②已知∠DOB=16∠AOB,∠BOE=23∠BOC,∠DOB与∠BOE互余,∴得:(...