$ \vec{ OA} + \vec{ OB} + \vec{ OC} + \vec{ OD}= 2 \vec{ OM} $ in circle

Question

Consider two perpendicular chords $AB$ and $CD$ of a given circle and ${M} = AB ∩ CD$. Show that $ \vec{ OA} + \vec{ OB} + \vec{ OC} + \vec{ OD}= 2 \vec{ OM} .$

I tried to write the vectors as : $$\vec{OA}=\vec{OM}+\vec{MA}, \vec{OB}=\vec{OM}+\vec{MB}, \vec{OC}=\vec{OM}+\vec{MC}, \vec{OD}=\vec{OM}+\vec{MD} $$ but it won't reduce when adding them. Can somebody give me another tips, please?

Answer 1

Let $P$ and $Q$ be mid-points of $CD$ and $AB$.

Thus, $$(\vec{ OA} + \vec{ OB}) + (\vec{ OC} + \vec{ OD})=2\vec{OQ}+2\vec{OP}= 2 \vec{ OM} $$