document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=174747>Question 174747</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The question is AOB, COD are right angles with OC in the interior of angleAOB and OB in the interior of angleCOD. OX is the bisector of angleAOC and OY is the bidsector of angleBOD. Can you prove that: angleAOX = angleBOY? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #129815 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=129815\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the measure of AOX. Then COX is also x. Since AOB is 90, then BOC = 90 - 2x. Since COD is also 90, then BOD = 90 - (90 - 2x) = 2x. Finally, BOY is half that, or x. Therefore, since AOX and BOY are both x, then AOX = BOY. <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=w%5E5 ALIGN=MIDDLE  ALT=\"w%5E5\"   DISABLED_style= \"max-width:90%; height:auto;\" >which was what we wanted </SPAN>\n" );
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