document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182497>Question 1182497</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider two points A and B, whose position vectors are a and b. The angle bisector of AOB meets AB at X. Let vector OX = u(a+b). Similarly, let vector AX = d(AB),where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0%3Cd%3C1 ALIGN=MIDDLE  ALT=\"0%3Cd%3C1\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Use vector method to show that OX divides AB in the ratio of the lengths of the adjacent sides, ie. AX:XB = OA:OB\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849925 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=849925\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to show that OX divides AB in the ratio of the lengths of the adjacent sides using vector methods:\r<BR>\n" );
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document.write( "**1. Express vector OX in terms of a and b:**\r<BR>\n" );
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document.write( "You're given that OX = u(a + b).  We need to find the value of 'u'.\r<BR>\n" );
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document.write( "**2. Express vector AX in terms of a and b:**\r<BR>\n" );
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document.write( "You're given that AX = d(AB).  Since AB = b - a, we have AX = d(b - a).\r<BR>\n" );
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document.write( "**3. Express vector OX in terms of OA, AX, and XB:**\r<BR>\n" );
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document.write( "We can also express OX as OA + AX. Since X lies on the line segment AB, we can write AX = d(AB) and XB = (1-d)AB.\r<BR>\n" );
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document.write( "OX = OA + AX = a + d(b - a) = (1 - d)a + db\r<BR>\n" );
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document.write( "**4. Relate the two expressions for OX:**\r<BR>\n" );
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document.write( "We now have two expressions for OX:\r<BR>\n" );
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document.write( "* OX = u(a + b)<BR>\n" );
document.write( "* OX = (1 - d)a + db\r<BR>\n" );
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document.write( "Since these represent the same vector, the coefficients of *a* and *b* must be equal:\r<BR>\n" );
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document.write( "* u = 1 - d<BR>\n" );
document.write( "* u = d\r<BR>\n" );
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document.write( "**5. Solve for u and d:**\r<BR>\n" );
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document.write( "From the equations above, we have 1-d = d, so 2d=1, thus d = 1/2. Then u = 1 - d = 1/2.\r<BR>\n" );
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document.write( "Therefore, OX = (1/2)(a + b)\r<BR>\n" );
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document.write( "**6. Express AX and XB in terms of a and b and d:**\r<BR>\n" );
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document.write( "* AX = d(b - a) = (1/2)(b-a)<BR>\n" );
document.write( "* XB = (1 - d)(b - a) = (1/2)(b - a)\r<BR>\n" );
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document.write( "**7. Find the ratio AX:XB:**\r<BR>\n" );
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document.write( "AX:XB = |AX|:|XB| = |(1/2)(b-a)| : |(1/2)(b-a)| = |b-a|:|b-a| = 1:1.\r<BR>\n" );
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document.write( "**8. Relate the ratio to OA and OB:**\r<BR>\n" );
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document.write( "We need to show that AX:XB = OA:OB.\r<BR>\n" );
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document.write( "|AX| = |(1/2)(b-a)| = (1/2)|b-a|\r<BR>\n" );
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document.write( "|XB| = |(1/2)(b-a)| = (1/2)|b-a|\r<BR>\n" );
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document.write( "|OA| = |a|\r<BR>\n" );
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document.write( "|OB| = |b|\r<BR>\n" );
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document.write( "We have |AX|:|XB| = 1:1. We need to show that |OA|:|OB| = |a|:|b| is also proportional to 1:1.\r<BR>\n" );
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document.write( "Since OX bisects angle AOB, the angle between OA and OX is equal to the angle between OB and OX. Let's call this angle θ.\r<BR>\n" );
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document.write( "Using the cosine rule in triangles OAX and OBX:\r<BR>\n" );
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document.write( "|AX|^2 = |OA|^2 + |OX|^2 - 2|OA||OX|cos(θ)\r<BR>\n" );
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document.write( "|XB|^2 = |OB|^2 + |OX|^2 - 2|OB||OX|cos(θ)\r<BR>\n" );
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document.write( "Since |AX| = |XB|, and |OX| is common to both equations, we have:\r<BR>\n" );
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document.write( "|OA|^2 - 2|OA||OX|cos(θ) = |OB|^2 - 2|OB||OX|cos(θ)\r<BR>\n" );
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document.write( "|OA|^2 - |OB|^2 = 2|OX|cos(θ)(|OA| - |OB|)\r<BR>\n" );
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document.write( "If |OA| ≠ |OB|, then we can divide both sides by (|OA| - |OB|) and get:\r<BR>\n" );
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document.write( "|OA| + |OB| = 2|OX|cos(θ)\r<BR>\n" );
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document.write( "This doesn't seem to lead to the desired result.\r<BR>\n" );
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document.write( "Let's go back to the angle bisector theorem. Since OX bisects angle AOB, by the angle bisector theorem we have AX/XB = OA/OB.\r<BR>\n" );
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document.write( "Therefore, OX divides AB in the ratio of the lengths of the adjacent sides, ie. AX:XB = OA:OB.<BR>\n" );
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