Geometry problem - CMI Entrance 2019

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Understand the problem

Let O be a point inside a parallelogram ABCD such that \(\angle AOB+\angle COD =180\) prove that \(\angle OBC =\angle ODC\)

Useful Points

C.M.I (Chennai mathematical institute UG-2019 entrance

Geometry 

5 out of 10

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Challenges and Thrills in Pre-College Mathematics

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This image has an empty alt attribute; its file name is Capture-300x141.png

Draw a clear image of the given problem.

Translate  ABCD along the vector AD SO A' and D are the same , and  so that B' and C are the same 

This image has an empty alt attribute; its file name is Capture2-1-300x131.png
This image has an empty alt attribute; its file name is Capture2-1-300x131.png

now , \(\angle COD +\angle CO'D\\=\angle COD+\angle A'O'D' \\=180 \)

so OCO'D is cyclic . therefore \(\angle OO'C =\angle ODC\)

 Also , vector BC and OO' both equal AD so OBCO' is parallelogram . therefore 

\(\angle OBC =\angle OO'C=\angle ODC \)

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