Question 1182025: A student tried to trisect an angle G using the following procedure:
1. Mark off GA congruent to GB.
2. Draw AB
3. Divide AB into 3 congruent parts so that AX=XY=YB.
4. Draw GX and GY.
Show that the student did not trisect angle G (Hint: Show that GA>GY. Then use indirect proof to show that m
I know how to show that GA>GY (by SAS Inequality) but cannot show why m
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! You're asking to prove that a student's method for trisecting an angle doesn't actually work. Here's a breakdown of the proof:
**1. Definitions and Setup**
* We're given that GA is congruent to GB, and AX = XY = YB. This means AY = (2/3)AB.
* We want to show that angle AGX, angle XGY, and angle YGB are *not* all equal.
**2. Showing GA > GY**
* Consider triangle GAY. By the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side. Therefore, GA + AY > GY.
* We know GA = GB. Since AX = XY = YB, Y lies between A and B, making AY < AB. So AY = (2/3)AB < AB.
* Now, assume, for the sake of contradiction, that GA <= GY. Since GA + AY > GY, then GY + AY > GY, which means AY > 0 (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.
**3. Indirect Proof**
* **Assumption:** Let's assume, for the sake of contradiction, that the student's method *does* trisect the angle. This means ∠AGX = ∠XGY = ∠YGB.
* **Consequence of the Assumption:** If the angles are equal, and we know 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.
* **Contradiction:** But we've already proven that GA > GY. If GX = GY, then we would also have GA = GY, which contradicts our earlier finding.
* **Conclusion:** Our initial assumption that the angles are equal must be false. Therefore, the student's method does *not* trisect the angle.
**In simpler terms:** The student's construction makes it look like the angle is trisected, but it's a visual trick. The side GY is shorter than GA, which prevents the equal angles from being created.
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