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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 $\angle OBC =\angle ODC$

##### Source of the problem
C.M.I (Chennai mathematical institute UG-2019 entrance
Geometry

5 out of 10

##### Suggested Book

Do you really need a hint? Try it first!

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

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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