document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=512636>Question 512636</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if ray ACFG and ray BCDE intersect at C and angle ADC is conngruent to angle BFC, then angle ADE is congruent to angle BFG.<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 #342686 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=342686\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "ADC and ADE are a linear pair and are therefore supplementary (the sum of their measures is 180).  Same relationship for BFC and BFG.\r<BR>\n" );
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document.write( "So mADC + mADE = 180\r<BR>\n" );
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document.write( "mBFC + mBFG = 180\r<BR>\n" );
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document.write( "Hence mADC + mADE = mBFC + mBFG\r<BR>\n" );
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document.write( "But we are given that mADC = mBFC, so substitute:\r<BR>\n" );
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document.write( "mBFC + mADE = mBFC + mBFG\r<BR>\n" );
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document.write( "Subtract mBFC from both sides\r<BR>\n" );
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document.write( "mADE = mBFG : <b>Q.E.D.</b>\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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