C

$\because BE$平分$\angle ABC$，
$\therefore \angle ABE=\angle EBF$，
$\because \angle BAC=90^{\circ}$，$\angle EFB=90^{\circ}$，
$\therefore \angle AEB=\angle BEF$，
$\because AD\bot BC$，$EF\bot BC$，
$\therefore AD$∥$EF$，$\angle AGE=\angle GEF$，
$\because \angle AEB=\angle BEF=\angle GEF$，
$\therefore \angle AGE=\angle AEB$，
$\therefore AG=AE$，可得①正确.
由①得$AG=AE\because \angle AEB=\angle BEF$，$\angle ABE=\angle EBF$，$BE=BE$，
$\therefore \triangle ABE$≌$\triangle FBE\left(ASA\right)$，
$\therefore AE=EF$，
$\therefore \angle EAF=\angle AFE$，
$\therefore \angle AQE=\angle FQE$，
$\because \angle AQE+\angle FQE=180^{\circ}$，
$\therefore \angle AQE=\angle FQE=90^{\circ}$，
$\because \triangle AGE$是等腰三角形，$AQ\bot GE$，
$\therefore GQ=QE$，
$\therefore {S_{△AGQ}}=\frac{1}{2}GQ⋅AQ=\frac{1}{2}QE⋅AQ={S_{△AQE}}$，可得②正确.
$\because AD$∥$EF$，
$\therefore \angle DAC=\angle FEC=\angle EAF+\angle EFA=2\angle EFA$，
$\because \angle EFA+\angle BEF=90^{\circ}$，$\angle EBC+\angle BEF=90^{\circ}$，
$\therefore \angle EFA=\angle EBC$，则$\angle DAC=2\angle EBC$，可得③正确
连接$GF$

$\because AQ=QF$，$\angle AQG=\angle GQF=90^{\circ}$，$GQ=GQ$，
$\therefore \triangle AGQ$≌$\triangle FGQ\left(SAS\right)$，
$\therefore AG=GF\because AG=AE=EF\therefore$四边形$AGFE$是菱形
要想$\triangle AGE$是等边三角形，则菱形$AGFE$中较小的角需要是$60^{\circ}$，
而题干中无法得知$\angle GAE$为$60^{\circ}$，故④不正确.
故选：$C$.