如图,$\triangle ABC$是直角三角形,$\angle ACB=90^{\circ}$,$CD\bot AB$于点$D$,$AF$是$\triangle ABC$的角平分线,过点$D$作$DG$∥$AF$交$BC$于点$G$,求证:$\angle CEF=\angle CGD$.请补全下面的证明过程.
证明:$\because CD\bot AB(已知)$,
$\therefore \angle ADC=90^{\circ}($______),
$\therefore \angle DAE+\angle AED=90^{\circ}(直角三角形两锐角互余)$,
$\because \angle ACB=90^{\circ}(已知)$,
$\therefore \angle \_\_\_\_\_\_+\angle CFA=90^{\circ}(直角三角形两锐角互余)$,
$\because AF$是$\triangle ABC$的角平分线（已知）,
$\therefore \angle CAF=\angle DAE($______),
$\therefore \angle AED=\angle CFA($______),
$\because \angle AED=\angle CEF($______),
$\therefore \angle CEF=\angle \_\_\_\_\_\_(等量代换)$,
$\because DG$∥$AF(已知)$,
$\therefore \angle CFA=\angle CGD($______),
$\therefore \angle CEF=\angle CGD( \_\_\_\_\_\_).$