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We know that: $$\angle ABO = \angle OBC$$ $$\angle DCO = \angle OCB$$ Prove: $$AB\parallel CD$$

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I have tried so many things: First I continued lines $BO$ and $CO$ to reach $CD$ and $AB$. Then I tried adding a parallel line to $AB$ that goes through point $O$. Also I tried adding a median to $BC$ that goes through point $O$, and in another attempt I tried it with an angle bisector. Also I have tried continuing lines $AB$ and $CD$ the other way and adding parallel lines to $BO$ and $OC$ which goes through points $C$ and $B$.

First I thought it seems so natural as just part of a Rhombus, but I found out that it isn't that easy as I thought it would be.

Please help me

Thanks

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    $\begingroup$ It isn't necessarily true. It is true if angle BOC is 90 degrees $\endgroup$ – David Quinn Feb 3 '17 at 18:06
  • $\begingroup$ Imagine moving $O$ up the page, so that $\alpha, \beta$ get close to 90°. Then the two side lines will be nowhere near parallel. $\endgroup$ – Joffan Feb 3 '17 at 18:49
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$$O_1=180-(\alpha+\beta)\\ \to o_2=180-o_1=\alpha+\beta\\$$now in $\triangle ODC$ we have $$o_2+D+c=180 \\D=180-(\alpha+\beta)-\beta$$ Note that $AB||CD $ only when $\hat{D}=\alpha$ so $$D=180-(\alpha+\beta)-\beta=\alpha \to 2(\alpha+\beta)=180 \\\to \alpha+\beta=90$$ This condition is necessary for $AB||CD $

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It looks like fake. Think about a regular hexagon with vertex $A,B,C,D$ (and others of course) and center $O$. Then all the angles are $60°$ but $AB,CD$ are not opposite edges, so no parallel

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