Answer:
[tex]GF=72[/tex]
Step-by-step explanation:
All triangles in the given figure are similar, from SAS. Notice that marked in the diagram, [tex]CD=DE=EF=FA[/tex].
For triangle [tex]\triangle CGF[/tex] was base [tex]GF[/tex], leg [tex]CF[/tex] contains three of these marked segments. In triangle [tex]\triangle CHE[/tex] with base [tex]HE[/tex], leg [tex]CE[/tex] has two of these marked segments. By definition, similar polygons have corresponding sides in a constant proportion. Therefore, the length of GF must be [tex]3/2[/tex] the length of EH. Since the length of EH is given as 48, we have:
[tex]GF=\frac{3}{2}EH, \\\\GF=\frac{3}{2}\cdot 48,\\\\GF=\boxed{72}[/tex]
