如图,射线$OM$平分$\angle BOA$.
$(1)$按要求尺规作图:作线段$AB$的垂直平分线交射线$OM$于点$C$,连接$CB$、$CA.(保留作图痕迹)$
$(2)$请把以下解题过程补充完整:
求证:$\angle OBC+\angle OAC=180^{\circ}$.
证明:在$OA$上截取$OD=OB$,连接$CD$.
$\because OM$平分$\angle BOA$,
$\therefore \angle BOM=\angle AOM$,
在$\triangle OBC$与$\triangle ODC$中:$\left\{\begin{array}{l}OB=OD\\∠BOM=∠AOM\\①\end{array}\right.$,
$\therefore \triangle OBC$≌$\triangle ODC($②______),
$\therefore CD=CB$,$\angle OBC=\angle ODC$,
$\because$点$C$在线段$AB$的垂直平分线上,
$\therefore CB=CA($③______),
$\therefore CD=CA$,
$\therefore \angle CDA=\angle CAD($④______),
$\because$点$D$在射线$OA$上,
$\therefore \angle ODC+\angle CDA=180^{\circ}$,
$\therefore \angle OBC+\angle OAC=180^{\circ}$.