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
