# Parallelogram and a point

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?

• AOD + BOC = 180 as well. ABC + ADC = 180... – Moti May 20 '19 at 6:52

Look at the picture:

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 $$CB\parallel OP$$.

Parallel translate the green triangle as shown in the figure:

$$\hspace{1cm}$$

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: $$\gamma =\angle OEC=\angle OBC$$.