document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182025>Question 1182025</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 m<XGY is not equal to m<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 m<XGY is not equal to m<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 #849956 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=849956\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 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:\r<BR>\n" );
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document.write( "**1. Definitions and Setup**\r<BR>\n" );
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document.write( "*   We're given that GA is congruent to GB, and AX = XY = YB. This means AY = (2/3)AB.<BR>\n" );
document.write( "*   We want to show that angle AGX, angle XGY, and angle YGB are *not* all equal.\r<BR>\n" );
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document.write( "**2. Showing GA > GY**\r<BR>\n" );
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document.write( "*   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.<BR>\n" );
document.write( "*   We know GA = GB. Since AX = XY = YB, Y lies between A and B, making AY < AB. So AY = (2/3)AB < AB.<BR>\n" );
document.write( "*   Now, assume, for the sake of contradiction, that GA <= GY. Since GA + AY > GY, then GY + AY > GY, which means AY > 0 (always true).<BR>\n" );
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.<BR>\n" );
document.write( "*   Since GA != GY and GA + AY > GY, then GA > GY.\r<BR>\n" );
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document.write( "**3. Indirect Proof**\r<BR>\n" );
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document.write( "*   **Assumption:** Let's assume, for the sake of contradiction, that the student's method *does* trisect the angle. This means ∠AGX = ∠XGY = ∠YGB.\r<BR>\n" );
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document.write( "*   **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.\r<BR>\n" );
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document.write( "*   **Contradiction:** But we've already proven that GA > GY. If GX = GY, then we would also have GA = GY, which contradicts our earlier finding.\r<BR>\n" );
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document.write( "*   **Conclusion:** Our initial assumption that the angles are equal must be false. Therefore, the student's method does *not* trisect the angle.\r<BR>\n" );
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document.write( "**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.<BR>\n" );
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