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(C) In $\Delta ABC$,$D$ and $E$ are the mid-points of $AB$ and $AC$ respectively. By the Mid-point Theorem,$DE \parallel BC$ and $DE = \frac{1}{2} BC$

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a parallelogram

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(C) In $\Delta ABC$,$D$ and $E$ are the mid-points of $AB$ and $AC$ respectively. By the Mid-point Theorem,$DE \parallel BC$ and $DE = \frac{1}{2} BC$.In $\Delta ABO$,$D$ and $P$ are the mid-points of $AB$ and $OB$ respectively. By the Mid-point Theorem,$DP \parallel AO$ and $DP = \frac{1}{2} AO$.In $\Delta ACO$,$E$ and $Q$ are the mid-points of $AC$ and $OC$ respectively. By the Mid-point Theorem,$EQ \parallel AO$ and $EQ = \frac{1}{2} AO$.From the above,we have $DP = EQ$ (both are equal to $\frac{1}{2} AO$) and $DP \parallel EQ$ (both are parallel to $AO$).Since one pair of opposite sides of the quadrilateral $DEQP$ is equal and parallel,$DEQP$ is a parallelogram.

$D$ and $E$ are the mid-points of the sides $AB$ and $AC$ of $\Delta ABC$ and $O$ is any point on side $BC$. $O$ is joined to $A$. If $P$ and $Q$ are the mid-points of $OB$ and $OC$ respectively,then $DEQP$ is

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