Question 1182174
<pre>
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.

{{{drawing(400,400,-4,4,-5,3,
line(-3,-4,3,-4), locate(-.07,2.3,G),
line(0,2,-3,-4),line(0,2,-1,-4), line(0,2,1,-4), line(0,2,3,-4),

locate(-3,-4,A),locate(-1,-4,X),locate(1,-4,Y),locate(3,-4,B) )}}}

△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

{{{drawing(400,400,-4,4,-5,3,
locate (-.1,1.1,k),locate(-.44,1.15,k),locate(.26,1.15,k),

line(-3,-4,3,-4), locate(-.07,2.3,G),
line(0,2,-3,-4),line(0,2,-1,-4), line(0,2,1,-4), line(0,2,3,-4),
green(line(-1,-4,-2.720294102,-3.440588204), locate(-3,-3.2,Z)),
locate(-3,-4,A),locate(-1,-4,X),locate(1,-4,Y),locate(3,-4,B) )}}}   

△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</pre>