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Let $ABCD$ be a parallelogram. $O$ is a point inside the parallelogram $ABCD$ such that $\angle AOB + \angle DOC = 180^\circ$. Prove that $\angle ODC= \angle OBC$. How can I prove it?

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  • $\begingroup$ AOD + BOC = 180 as well. ABC + ADC = 180... $\endgroup$ – Moti May 20 '19 at 6:52
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Look at the picture:

enter image description here

Consider another copy of the parallelogram, i.e. consider the translation for vector $\vec{DA}$. Then $D$ maps to $A$, $C$ maps to $B$, and let $A$ maps to $F$, $B$ to $E$, and $O$ to $P$. Since $180^\circ= \angle AOB+\angle COD= \angle AOB+\angle BPA$ we conclude that $\square APBO$ is cyclic. So $\angle ODC= \angle PAB= \angle POB= \angle OBC$, where the second equality holds by the equality of inscribed angles over $BP$, and the third one as $CB\parallel OP$.

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Parallel translate the green triangle as shown in the figure:

$\hspace{1cm}$enter image description here

Since $\alpha+\beta=180^\circ$, the quadrilateral $OCED$ is cyclic.

In the circle, the inscribed angles subtending the same arc are equal: $\gamma =\angle ODC=\angle OEC$.

Finally, the quadrilateral $OBCE$ is a parallelogram, hence, the opposite angles are equal: $\gamma =\angle OEC=\angle OBC$.

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