结论：$BD=CE$，$BD\bot CE$，
理由：如图，设$AB$，$BD$与$CE$交于点$O$，$Q$，
$\because AB\bot AC$，$AD\bot AE$，
$\therefore \angle BAC=\angle DAE=90^{\circ}$，
$\therefore \angle BAD=\angle CAE$，
在$\triangle ABD$和$\triangle ACE$中，
$\left\{\begin{array}{l}{AB=AC}\\{∠BAD=∠EAC}\\{AD=AE}\end{array}\right.$，
$\therefore \triangle ABD$≌$\triangle ACE\left(SAS\right)$，
$\therefore \angle ABD=\angle ACE$，$BD=CE$，
$\because \angle ACO+\angle AOC=90^{\circ}$，
$\therefore \angle OBQ+\angle BOQ=90^{\circ}$，
$\therefore \angle BQO=90^{\circ}$，
$\therefore BD\bot CE$.

结论：$BD=CE$，$BD\bot CE$，
理由：如图，设$AB$，$BD$与$CE$交于点$O$，$Q$，
$\because AB\bot AC$，$AD\bot AE$，
$\therefore \angle BAC=\angle DAE=90^{\circ}$，
$\therefore \angle BAD=\angle CAE$，
在$\triangle ABD$和$\triangle ACE$中，
$\left\{\begin{array}{l}{AB=AC}\\{∠BAD=∠EAC}\\{AD=AE}\end{array}\right.$，
$\therefore \triangle ABD$≌$\triangle ACE\left(SAS\right)$，
$\therefore \angle ABD=\angle ACE$，$BD=CE$，
$\because \angle ACO+\angle AOC=90^{\circ}$，
$\therefore \angle OBQ+\angle BOQ=90^{\circ}$，
$\therefore \angle BQO=90^{\circ}$，
$\therefore BD\bot CE$.