如图,在$\triangle ABC$中,点$D$、$E$在边$AB$上,点$F$、$G$分别在边$BC$、$AC$上,$\angle ACB=\angle BEC=\angle BDF=90^{\circ}$,$\angle GEC+\angle CFD=180^{\circ}$,试说明$EG\bot AC$.请完善解答过程,并在括号内填写相应的理论依据.
解:$\angle BEC=\angle BDF=90^{\circ}$,（已知）
$\therefore CE$∥$\_\_\_\_\_\_,\left( \_\_\_\_\_\_\right)$
$\therefore \angle ECB+\angle CFD=180^{\circ}.($______)
$\because \angle GEC+\angle CFD=180^{\circ}$,（已知）
$\therefore \angle GEC=\angle ECB.($______)
$\therefore GE$∥$BC.\left( \_\_\_\_\_\_\right)$
$\therefore \angle AGE=\angle ACB=90^{\circ}.($______)
$\therefore EG\bot AC.($______)