document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=199732>Question 199732</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two chords of a circle are perpendicular and congruent, does one of them have a diameter? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #150100 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=150100\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Very  interesting  question<BR>\n" );
document.write( "Let's  start  with  a  rough  sketch  of  a  circle,  now  one  chord,  now  another  perpendicular  chord  that  goes  thru  center.<BR>\n" );
document.write( "This  is  the  situation  that  answers  your  question,  yes  it  can  happen  but  it  does  not have  to.<BR>\n" );
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document.write( "The  principle  that  puts  the  second  chord  thru  the  center  is,,,, it  must  be  a  perpendicular  bisector. ALL  perp  bisectors  go  thru center,  hence  could  be  diameters.<BR>\n" );
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document.write( "It  is  easy  to  take  your  sketch,  and  just  slide  the  second  chord to  one  side  or  the  other. It   does  not  go  thru  center,  and  is  not  a  diameter.  Your  problem  did  not  specify  bisection.<BR>\n" );
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document.write( "Another  interesting  construction,  again  draw  a  rough  circle  and  one  chord.  Draw  a  second  chord, perpendicular  to  first  but  located  at  an  end  of  the  first  chord.  NOW,  if  we  connect  the  other  ends of  the  chord,  we  have  a  triangle.   But,  a  very  special  triangle, it  is  a  right inscribed  triangle,  that  divides  the  circle  into  two  semicircles,  and  has  the  hypotenuse  as  the  diameter.\r<BR>\n" );
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document.write( "Good  luck  with  class<BR>\n" );
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