ABCD is a parallelogram in which diagonals AC | vedclass

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(C) In $\Delta OAB$,$E$ is the midpoint of $AO$ and $H$ is the midpoint of $BO$. By the Midpoint Theorem,$EH = \frac{1}{2} AB$.Similarly,in $\Delta OB

Vedclass · Accepted Answer

$1:2$

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(C) In $\Delta OAB$,$E$ is the midpoint of $AO$ and $H$ is the midpoint of $BO$. By the Midpoint Theorem,$EH = \frac{1}{2} AB$.Similarly,in $\Delta OBC$,$F$ is the midpoint of $DO$ and $G$ is the midpoint of $CO$. Note that the vertices are labeled such that $EFGH$ forms a quadrilateral inside. Applying the Midpoint Theorem to triangles $\Delta OAB, \Delta OBC, \Delta OCD$,and $\Delta ODA$:$EH = \frac{1}{2} AB$$HG = \frac{1}{2} CD$$GF = \frac{1}{2} BC$$FE = \frac{1}{2} DA$Perimeter of $EFGH = EH + HG + GF + FE = \frac{1}{2} (AB + CD + BC + DA) = \frac{1}{2} (	ext{Perimeter of } ABCD)$.Therefore,the ratio of the perimeter of $EFGH$ to the perimeter of $ABCD$ is $\frac{1}{2} : 1$,which is $1:2$.

$ABCD$ is a parallelogram in which diagonals $AC$ and $BD$ intersect at $O$. If $E, F, G$ and $H$ are the midpoints of $AO, DO, CO$ and $BO$ respectively,then the ratio of the perimeter of the quadrilateral $EFGH$ to the perimeter of parallelogram $ABCD$ is

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