Question 1182026
Here's how to show the student's method doesn't trisect the angle:

1. **GA > GY:**

   Consider triangle GAY.  We know GA = GB (by construction) and AX = XY = YB (also by construction).  Therefore, AY = 2/3 AB.  Since GA = GB, triangle GAB is isosceles.

   Now, in triangle GAY, GA and AY are two sides. The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side. Therefore, GA + AY > GY.

   Also, in triangle GAB, since GA = GB and AX = XY = YB, Y is between A and B, which means that AY < AB. Hence AY = (2/3)AB < AB. 

   If we assume that GA <= GY, then since GA + AY > GY, we have GY + AY > GY. This means AY > 0, which is always true. 

   However, if we assume GA = GY, then triangle GAY is isosceles. Since A, X, and Y are collinear, this would imply that GX is the angle bisector of angle AGY. In triangle GAB, since GA = GB, the median from G to AB will bisect angle AGB. Since GX is not a median, GX is not the angle bisector. Therefore, GA != GY.

   Since GA != GY and GA + AY > GY, then GA > GY.

2. **Indirect Proof that ∠XGY ≠ ∠XGA:**

   * **Assumption:** Let's assume, for the sake of contradiction, that ∠XGY = ∠XGA.

   * **Consequence of the Assumption:** If ∠XGY = ∠XGA, and we already know that GA = GB and AX = XY = YB, then by the Side-Angle-Side (SAS) congruence theorem, triangles GAX and GYX would be congruent.  This would imply that GX = GY.  Furthermore, if ∠XGY = ∠XGA, then since ∠AGB = ∠AGX + ∠XGY + ∠YGB, and if ∠XGY = ∠XGA, then this would imply that GX is the angle bisector of ∠AGB.

   * **Contradiction:** But we have already shown that GA > GY. If GX = GY, then GA = GY. This contradicts the fact that GA > GY. Therefore, our assumption that ∠XGY = ∠XGA must be false.

   * **Conclusion:** Therefore, ∠XGY ≠ ∠XGA. The student's method does *not* trisect angle G.