I am like you and can see no way that the hint is any good.  But rather than
leave you hanging, I'll prove it another way, but also using indirect proof.



△AGB is isosceles GA and GB are congruent

It's easy to prove that △GXY is isosceles.

△GAX and △GBY are congruent by SAS, so GX = GY by CPCT.

Let's assume (for contradiction) that ∠G is trisected and each third
of ∠G equals, say k.

Locate point Z such that GZ = GX
Draw in XZ

   

△ZGX ≅ △XGY  by SAS

XZ = XY   by CPCT

XY = AX   given

XZ = AX   transitive axiom ("things equal to the same thing are equal")   

△XAZ is isosceles.

∠AZX is acute because base angles of any isosceles triangle are acute.
      
△GZX is isosceles.

∠GZX is acute because base angles of any isosceles triangle are acute.

∠AZX and ∠GZX are supplementary (they form a linear pair)

We have reached a contradiction because two acute angles cannot be
supplementary.

Therefore the assumption that the student trisected angle G is false,
so he or she did not trisect the angle.

Edwin