document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=455308>Question 455308</A>: <I><SPAN STYLE=\"word-break: break-all;\"> R, S, and T are, respectively, the midpoints of the sides AB, AD, and CD of a square. Prove that DR and ST trisect each other. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #312938 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=312938\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C0%2C10%2C0%2C10%2C%0D%0A%0D%0A%0D%0Aline%282%2C2%2C8%2C2%29%2C%0D%0Aline%288%2C2%2C8%2C8%29%2C%0D%0Aline%288%2C8%2C2%2C8%29%2C%0D%0Aline%282%2C8%2C2%2C2%29%2C%0D%0Alocate%282%2C2%2CA%29%2C%0D%0Alocate%288%2C2%2CB%29%2C%0D%0Alocate%288%2C8%2CC%29%2C%0D%0Alocate%282%2C8%2CD%29%2C%0D%0Alocate%285%2C2%2CR%29%2C%0D%0Alocate%282%2C5%2CS%29%2C%0D%0Alocate%285%2C8%2CT%29%2C%0D%0Aline%282%2C8%2C5%2C2%29%2C%0D%0Aline%282%2C5%2C5%2C8%29%2C%0D%0Aline%285%2C2%2C5%2C8%29%0D%0A%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "I presume your question is \"Prove that DR and ST split the other segment in a ratio of 1:2\" (since \"trisect means \"to split into three equal parts\").\r<BR>\n" );
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document.write( "Let P be the intersection of DR and ST. One can show, using AAA similarity, that triangles SPD and TPR are similar. Also, TR:SD = 2:1, so we can let TR = s and SD = s/2. Since the ratios among similar triangles are equal, this implies that TP:SP = 2:1 and RP:QP = 2:1. This means that we can let TP = 2k, SP = k for some k and note that TP/TS = TP/(TP + SP) = 2k/(2k + k) = 2/3, and the same symmetry applies to the other lengths. </SPAN>\n" );
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