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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

challenges and thrills of pre college mathemetics

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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