document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1005149>Question 1005149</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if in  ABC triangle &#8736;C=60&#8728; then prove that 1/(a+c)+1/(b+c)=3/(a+b+c) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #621513 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=621513\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Multiply both sides by (a+c)(b+c)(a+b+c) to clear out the fractions and make life easier. That is, statement holds if and only if\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large (b+c)(a+b+c) + (a+c)(a+b+c) = 3(a+c)(b+c)\">\r<BR>\n" );
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document.write( "With some simplification, this is equivalent to\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large a^2 + b^2 = c^2 + ab\">\r<BR>\n" );
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document.write( "or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large c^2 = a^2 + b^2 - ab\">\r<BR>\n" );
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document.write( "However this statement is true by applying the law of cosines on triangle ABC, as angle C = 60 degrees. </SPAN>\n" );
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