$(1)证明:∵CD⊥OM,CE⊥ON$ $∴∠CDA=∠CEB=∠CDO=90°$ $在Rt△ADC和Rt△BEC中$
${{\begin{cases}{{AC=BC}}\\{AD=BE}\end{cases}}} $ $∴Rt△ADC≌Rt△BEC(HL)$ $∴CD=CE$ $又∵CD⊥OM,CE⊥ON$ $∴OC平分∠MON$ $(2)(更多请点击查看作业精灵详解)$
$5.(2)解:在Rt△OCD和Rt△OCE中$ $\begin{cases}{OC=OC}\\{CD=CE}\end{cases}$ $∴Rt△OCD≌Rt△OCE(HL)$ $∴OD=OE\ $ $∵AD=3,BO=4,AD=BE\ $ $∴OE=OB+BE=4+3=7=OD\ $ $∴OA=OD+AD=7+3=10$
|
|