The cool thing about cyclic quadrilaterals is that the properties they hold often work in reverse. For example, it's easy to prove that if $ABCD$ is cyclic, then e.g. $\angle ABC+\angle ADC=180^\circ$. However, it turns out the reverse is true as well: if $\angle ABC+\angle ADC=180^\circ$, then $ABCD$ is cyclic. This is immensely powerful when put to good use!
And now, the solution to the original problem. I claim that quadrilateral $GDFC$ is cyclic. To see this, remark that $$\angle GDF+\angle GCF=90^\circ+90^\circ=180^\circ,$$ so the opposite angles of the quadrilateral are supplementary, meaning the quadrilateral is cyclic as desired.
Now a simple angle chase yields $$\angle BAC=\angle BDC\equiv\angle GDC=\angle GFC$$ which is what we wanted.