Question 1182026: 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 angle XGY is not equal to angle XGA.)
I know how to show that GA>GY (by SAS Inequality) but cannot show why angle XGY is not equal to angle XGA using indirect proof. I appreciate any help! Thanks in advance!
Found 2 solutions by CPhill, Edwin McCravy:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
if two sides of one triangle are congruent to two sides of another triangle, then
the third side of the triangle with the larger included angle is longer.
GA congruent to GB
angle A congruent to angle B
AX congruent to YB
GX congruent to GY
triangle XGY is isosceles
GX congruent to GY
AX congruent to XY
angles GXY and GYX are acute (base angles of an isosceles triangle)
angle GXA is obtuse (supplementary to an acute angle)
GA > GX (by the SAS inequality theorem.
GA > GY (since GX is congruent to GY <-----------STEP M
For contradiction, assume angle XGY congruent to angle XGA
Extend GX to twice its length to E such that GX congruent to EX.
Draw YE
AX congruent to XY (given)
GX congruent to EX (by construction)
angle GXA congruent to angle EXY (vertical angles are congruent)
triangle GAX and triangle EYX are congruent by SAS
GA congruent to EY (c.p.c.t) <----------------- STEP N
angle XGA congruent to angle YEX (c.p.c.t.)
angle YEX congruent to angle XGY
Triangle GYE is isosceles (base angles congruent)
GY congruent to EY (c.p.c.t) <------------------ STEP P
From steps N and P above,
GA is congruent to GY <----------------- Step Q
Step Q contradicts Step M.
Therefore, the assumption that angle XGY congruent to angle XGA
is false, so the student did not trisect angle AGB.
Edwin
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