SOLUTION: 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

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 $EF$.
Answer by KMST(5336) (Show Source): You can put this solution on YOUR website!
The data does not add up. I cannot draw any parallel lines between those points.
If sides EF and GH are parallel, The measures of arcs FG and HE must be the same.
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.
AS POSTED:
If a quadrilateral inscribed in a circle is called
it means that going around the circle in one direction we find the points , , , and one right after the other in that order.
That means that going from towards we passed through .
Then , or , so --> .
The of arc include the of arc plus the measure of arc .
Then
So, -->
,
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