$\bullet$ $CA=CB:$ Let $E$ and $F$ be midpoints of $AD$ and $AB$,respectively. Since $GE||BC$ we get $F-E-G$ are collinear $\implies AF=FB=FD$. $\angle GCA=\angle CAB=\angle CBA=\angle GFA \implies GCAF$ is cyclic $\implies \angle AGC=180-\angle CFA=180-90=90. \square$ $\bullet$ $\angle AGC=90:$ $AGCD$ is cyclic. Let $\angle AGE=\angle DGE=\angle GDC=\alpha \implies CAD=\angle CGD=180-\alpha-\angle GCD=180-\alpha-(180-\angle GAD)=90-2\alpha \implies \angle GAC=\alpha \implies \angle DAF=\alpha \implies \angle CBA=\angle CAB=90-\alpha \implies CA=CB. \blacksquare$

see the figure:

Attachments:

$GM\parallel BC, AB\parallel BC$, implies $AMCG$ is a parallelogram. $\angle AGC=90^\circ\Leftrightarrow \angle AMC=90^\circ\Leftrightarrow AC=BC$ since $M$ is midpoint of $AB$.

Let $M$ be the midpoint of $\overline{AB}$ and note $BCGM$ is a parallelogram. Then, $MD=MB=GC$ so $CDMG$ is a cyclic isosceles trapezoid. Suppose $\angle AGC=90.$ Then, $ADCG$ and therefore $AMDCG$ is cyclic so $\angle MAC=\angle MGC=\angle CBM$ and $AC=BC.$ Conversely, if $AC=BC,$ then $\angle MAC=\angle ABC=\angle MGC$ so $AMDCG$ is cyclic and $\angle AGC=90.$ $\square$