根据题意,完成作图及推理:
已知:在$\triangle ABC$中,$\angle C=90^{\circ}$,$\angle BAC$的平分线交$BC$于点$D$,$CD=1$,$AB=4$.
$(1)$作出$\angle BAC$的平分线（尺规作图,保留作图痕迹）；
$(2)$求$\triangle ABD$的面积.
解:过$D$作$DE\bot AB$于点$E$,
$\because \angle C=90^{\circ}$,
$\therefore DC\bot$______于点$C$,
$\because AD$平分$\angle BAC$,
$\therefore \_\_\_\_\_\_=CD( \_\_\_\_\_\_)$,
$\because CD=1$,
$\therefore DE=$______,
$\because AB=4$,
$\therefore S_{\triangle ABD}= \_\_\_\_\_\_.$