document.write( "Question 1210566: Trapezoid $HGFE$ is inscribed in a circle, with $\overline{EF} \parallel \overline{GH}$. If arc $EG$ is $40$ degrees, arc $EH$ is $120$ degrees, and arc $FG$ is $20$ degrees, find arc $EF$. \n" ); document.write( "

Algebra.Com's Answer #853704 by KMST(5336) $\"\"$ $\"About$

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The data does not add up. I cannot draw any parallel lines between those points.
\n" ); document.write( "If sides EF and GH are parallel, The measures of arcs FG and HE must be the same.
\n" ); document.write( "Maybe EF and EG were both supposed to measure 120 degrees, and 40 degrees was the measure of arc HG instead of being the measure of EG.
\n" ); document.write( "AS POSTED:
\n" ); document.write( "If a quadrilateral inscribed in a circle is called $\"HGFE\"$
\n" ); document.write( "it means that going around the circle in one direction we find the points $\"H\"$ , $\"G\"$ , $\"F\"$ , and $\"E\"$ one right after the other in that order.
\n" ); document.write( "That means that going from $\"G\"$ towards $\"E\"$ we passed through $\"F\"$ .
\n" ); document.write( "Then $\"GF%2BFE=GE\"$ , or $\"FG%2BEF=EG\"$ , so $\"20%5Eo%2BEF=%2440%5Eo\"$ --> $\"EF=40%5Eo-20%5Eo=20%5Eo\"$ .
\n" ); document.write( "The $\"40%5Eo\"$ of arc $\"EG\"$ include the $\"20%5E0\"$ of arc $\"FG\"$ plus the measure of arc $\"EF\"$ .
\n" ); document.write( "Then $\"HG%2BGF%2BFE%2BEH=HG%2B20%5Eo%2B20%5Eo%2B120%5Eo=360%5E0\"$
\n" ); document.write( "So, $\"HG%2B160%5Eo=360%5Eo\"$ --> $\"HG=360%5Eo-160%5Eo=200%5Eo\"$
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