Since $\overline{OA} \parallel \overline{BC}$, we have that
So I know that $\angle COA=\angle BCO=40^\circ$. I'm trying to figure out how to make use of the facts that these triangles are inscribed in circles… help appreciated!
Finding an angle involving two triangles inscribed in a circle
euclidean-geometrygeometryplane-geometry
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Hint:
Sides $c$ and $a$ are given, and angle $A$. Notice that there are two possible triangles with side length $a$, and angle $C'$ is the supplement of $C$. Hopes this helps.
I think I have a correct proof now.
Proof by contradiction: Assume to the contrary that two lines parallel to the same line are not parallel to each other. Without loss of generality, assume line m and line n are parallel to a line l, but m and n are not parallel to each other. Then, m and n intersect at a point, P that is not on line l. However, this contradicts Axiom 5 because two lines would be containing P and be parallel to l. So the assumption that m and n are not parallel was incorrect. Thus, m and n are parallel to l and also parallel to each other.
Best Answer
$\angle O =\angle OCB = 40^°$ (alternate interior angles, $OA \parallel BC$)
$\angle B =\frac{\angle O}{2} = 20^°$ ($\angle$ of center=2$\angle$ circumference subtended same segment)
$\angle ODA=120^°$(sum of $\angle$ 's in a $\triangle$)