Answer: 32
Step-by-step explanation:
Since [tex]\overline{FG} \parallel \overline{AC}[/tex], we know that by the corresponding angles theorem, [tex]\angle BFG \cong \angle BAC[/tex] and [tex]\angle BGF \cong \angle BCA[/tex].
Thus, [tex]\triangle BFG \sim \triangle BAC[/tex] by AA.
It follows that since corresponding sides of similar triangles are proportional,
[tex]\frac{36+90}{90}=\frac{CG+80}{80}\\ \\\frac{126}{90}=\frac{CG+80}{80}\\\left(\frac{126}{90} \right)(80)=CG+80\\112=CG+80\\CG=\boxed{32}[/tex]
