document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=720830>Question 720830</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello there, I am hoping that you can help me with this geometry proof. I have been looking everywhere for an example, but have had no luck. I even checked geometry books out at our local library.\r<BR>\n" );
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document.write( "I believe that the problem is based on the Pythagorean Theorem Proof Using Similarity. \r<BR>\n" );
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document.write( "Given: &#8710; ABC, AD bisects &#8736;BAC, and AE &#8773; ED			       <BR>\n" );
document.write( "Prove: AE/AC = BD/BC<BR>\n" );
document.write( "The picture provided shows a triangle labeled ABC. AD bisects angle BAC. From point D there is another line that extends to the side of the triangle labeled AC, this point is labeled E. I am sorry that I am unable to attach a picture. Any help would be appreciated!<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 #442005 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=442005\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You have to prove angle DEC is congruent to angle BAC <BR>\n" );
document.write( "( exterior angle theorem) <BR>\n" );
document.write( "angle DEC = angle DAE +EDA\r<BR>\n" );
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document.write( "But DAE =ADE=BAD ( given)\r<BR>\n" );
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document.write( "In triangles ABC & CED angle C is common angle.<BR>\n" );
document.write( "DEC = BAC\r<BR>\n" );
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document.write( "Triangles are similar.\r<BR>\n" );
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document.write( "Take the ratio of sides ( basic proportionality theorem)\r<BR>\n" );
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document.write( "EC/AC = DC/BC\r<BR>\n" );
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document.write( "(AC-AE)/AC = (BC-BD)/BC\r<BR>\n" );
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document.write( "AC/AC  - AE/AC = BC/BC- BD/BC\r<BR>\n" );
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document.write( "1-AE/AC = 1- BD/BC\r<BR>\n" );
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document.write( "therefore AE/AC = BD/BC\r<BR>\n" );
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