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

Algebra ->  Geometry-proofs -> 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      Log On


   

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) About Me  (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 HGFE
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.
That means that going from G towards E we passed through F.
Then GF%2BFE=GE, or FG%2BEF=EG, so 20%5Eo%2BEF=%2440%5Eo --> EF=40%5Eo-20%5Eo=20%5Eo .
The 40%5Eo of arc EG include the 20%5E0 of arc FG plus the measure of arc EF.
Then HG%2BGF%2BFE%2BEH=HG%2B20%5Eo%2B20%5Eo%2B120%5Eo=360%5E0
So, HG%2B160%5Eo=360%5Eo --> HG=360%5Eo-160%5Eo=200%5Eo
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