Let O be a point inside a parallelogram ABCD such that \(\angle AOB+\angle COD =180\) prove that \(\angle OBC =\angle ODC\)
C.M.I (Chennai mathematical institute UG-2019 entrance
Geometry
5 out of 10
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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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