document.write( "Question 806126: Given: Quadrilateral ABCD is a parallelogram.Segment DE is perpendicular to Segment AC, Segment GF is perpendicular to Segment AC.
\n" ); document.write( "Prove: AE plus CF all over CF is equal to ED plus FG all over FG\r
\n" ); document.write( "\n" ); document.write( "Please!!! TOTALLY NEED THE ANSWER!!!! \n" ); document.write( "

Algebra.Com's Answer #485703 by mananth(16946) $\"\"$ $\"About$

You can put this solution on YOUR website!
DE is perpendicular to AC
\n" ); document.write( "GF is perpendicular to AX\r
\n" ); document.write( "\n" ); document.write( "so DE is parallel to GF\r
\n" ); document.write( "\n" ); document.write( "In triangle DEC & GFC\r
\n" ); document.write( "\n" ); document.write( "angle DEC is congruent to triangle GFC
\n" ); document.write( "Angle EDC is congruent to FGC ( corresponding angles)\r
\n" ); document.write( "\n" ); document.write( "so the triangles are similar by AA test of similarity\r
\n" ); document.write( "\n" ); document.write( "Therefore EC/FC = ED/FG = DC/GC ratios of corresponding sides are equal\r
\n" ); document.write( "\n" ); document.write( "But EC = AC ( diagonals of a parallelogram bisect each other)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Therefore (AC+FC)/FC = (ED+FG)/FG ( by componendo)\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );