document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182026>Question 1182026</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A student tried to trisect an angle G using the following procedure:<BR>\n" );
document.write( "1. Mark off GA congruent to GB.<BR>\n" );
document.write( "2. Draw AB<BR>\n" );
document.write( "3. Divide AB into 3 congruent parts so that AX=XY=YB.<BR>\n" );
document.write( "4. Draw GX and GY.<BR>\n" );
document.write( "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.)\r<BR>\n" );
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document.write( "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! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849955 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849955\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to show the student's method doesn't trisect the angle:\r<BR>\n" );
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document.write( "1. **GA > GY:**\r<BR>\n" );
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document.write( "   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.\r<BR>\n" );
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document.write( "   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.\r<BR>\n" );
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document.write( "   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. \r<BR>\n" );
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document.write( "   If we assume that GA <= GY, then since GA + AY > GY, we have GY + AY > GY. This means AY > 0, which is always true. \r<BR>\n" );
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document.write( "   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.\r<BR>\n" );
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document.write( "   Since GA != GY and GA + AY > GY, then GA > GY.\r<BR>\n" );
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document.write( "2. **Indirect Proof that ∠XGY ≠ ∠XGA:**\r<BR>\n" );
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document.write( "   * **Assumption:** Let's assume, for the sake of contradiction, that ∠XGY = ∠XGA.\r<BR>\n" );
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document.write( "   * **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.\r<BR>\n" );
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document.write( "   * **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.\r<BR>\n" );
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document.write( "   * **Conclusion:** Therefore, ∠XGY ≠ ∠XGA. The student's method does *not* trisect angle G.<BR>\n" );
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