document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=762959>Question 762959</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given: AX || CY<BR>\n" );
document.write( "BA bisects &#8736;CAX, <BR>\n" );
document.write( "BC bisects &#8736;ACY.<BR>\n" );
document.write( "Prove: &#8736;B is a right angle.\r<BR>\n" );
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document.write( "Diagram looks like a triangle (where it's pointed end is pointing in the right) in the middle of the parallel lines. \r<BR>\n" );
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document.write( "Looks something like this:\r<BR>\n" );
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document.write( "A<BR>\n" );
document.write( "__________X<BR>\n" );
document.write( "|<BR>\n" );
document.write( "|   .C<BR>\n" );
document.write( "|_________Y<BR>\n" );
document.write( "B\r<BR>\n" );
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document.write( "Has to be written as Statements & Reasons. I really want to know how to do this! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #464610 by </B><A HREF=/tutors/aboutme.mpl?userid=ramkikk66><B>ramkikk66(644)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ramkikk66><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=464610\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "See the picture below which describes your problem. I have extended line CA further upwards to a point P, which is required for the proof.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing+%28600%2C600%2C-10%2C10%2C-20%2C20%2C+%0D%0Aline%28-2%2C3.5%2C8%2C3.5%29%2C%0D%0Aline%28-1%2C9%2C8%2C9%29%2C%0D%0Aline%28-2%2C3.5%2C-0.55%2C12%29%2C%0D%0Aline%28-2%2C3.5%2C0.5%2C6.5%29%2C%0D%0Aline%28-1%2C9%2C0.5%2C6%2C5%29%2C%0D%0Alocate%280.5%2C6.5%2CB%29%2C%0D%0Alocate%28-1%2C9%2CA%29%2C%0D%0Alocate%288%2C9%2CX%29%2C%0D%0Alocate%28-2%2C3.5%2CC%29%2C%0D%0Alocate%288%2C3.5%2CY%29%2C%0D%0Alocate%28-0.55%2C12%2CP%29%0D%0A%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Problem:<BR>\n" );
document.write( "AX || CY<BR>\n" );
document.write( "BA bisects ?CAX,<BR>\n" );
document.write( "BC bisects ?ACY.<BR>\n" );
document.write( "Prove: ?B is a right angle. \r<BR>\n" );
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document.write( "Step 1:<BR>\n" );
document.write( "BA bisects angle CAX. So angle CAB = angle BAX\r<BR>\n" );
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document.write( "Step 2:<BR>\n" );
document.write( "BC bisects angle ACY. So angle ACB = angle BCY\r<BR>\n" );
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document.write( "Step 3:<BR>\n" );
document.write( "Angle PAX = angle ACY since AX || CY (external angles of parallel lines)\r<BR>\n" );
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document.write( "Step 4:<BR>\n" );
document.write( "Angles PAX + CAX = 180 (supplementary angles on a straight line)\r<BR>\n" );
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document.write( "Step 5:<BR>\n" );
document.write( "i.e. PAX / 2 + CAX / 2 = 90\r<BR>\n" );
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document.write( "Step 6:<BR>\n" );
document.write( "ACY / 2 + CAX / 2 = 90  Since PAX = ACY\r<BR>\n" );
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document.write( "Step 7:<BR>\n" );
document.write( "ACB + CAB = 90  Since ACB is half of ACY, CAB is half of CAX\r<BR>\n" );
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document.write( "Step 8:<BR>\n" );
document.write( "But in Triangle ABC, ACB + CAB + ABC = 180\r<BR>\n" );
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document.write( "Step 9:<BR>\n" );
document.write( "So ABC = Angle B = 90\r<BR>\n" );
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document.write( ":)\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );