$ABCD$ is a parallelogram and $O$ is any point. The parallelogram $OAEB,OBFC,OCGD,ODHA$ are completed. show that $EFGH$ is a parallelogram.
2026-03-27 16:20:55.1774628455
Show that EFGH is a parallelogram
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If you say $O$ is the origin of position vectors, then you have $$E = A+B\;\;\;{\rm and} \;\;\;F = B+C\;\;\;{\rm and} \;\;\;G = C+D\;\;\;{\rm and} \;\;\; H=A+D $$ Now $$EF = F-E = C-A = G-H = HG$$ so $EFGH$ is a parallelogram and you don't need that $ABCD$ is a parallelogram.