In $\Delta ABC$,$P, Q,$ and $R$ are the midpoints of $AB, BC,$ and $CA$ respectively. Prove that $PBQR, PQCR,$ and $PQRA$ all are parallelograms.
(N/A) Given: $P, Q,$ and $R$ are midpoints of sides $AB, BC,$ and $CA$ of $\Delta ABC$ respectively.
By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.
$1$. Consider $PBQR$: Since $R$ is the midpoint of $AC$ and $P$ is the midpoint of $AB$,$PR \parallel BC$ and $PR = \frac{1}{2} BC = BQ$. Also,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = RC$. Since $PR \parallel BQ$ and $PQ \parallel RB$,$PBQR$ is a parallelogram.
$2$. Consider $PQCR$: Since $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = RC$. Also,$QR \parallel AB$ and $QR = \frac{1}{2} AB = PC$. Since $PQ \parallel RC$ and $QR \parallel PC$,$PQCR$ is a parallelogram.
$3$. Consider $PQRA$: Since $Q$ is the midpoint of $BC$ and $R$ is the midpoint of $AC$,$QR \parallel AB$ and $QR = \frac{1}{2} AB = AP$. Also,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = AR$. Since $QR \parallel AP$ and $PQ \parallel AR$,$PQRA$ is a parallelogram.
By the Midpoint Theorem,the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.
$1$. Consider $PBQR$: Since $R$ is the midpoint of $AC$ and $P$ is the midpoint of $AB$,$PR \parallel BC$ and $PR = \frac{1}{2} BC = BQ$. Also,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = RC$. Since $PR \parallel BQ$ and $PQ \parallel RB$,$PBQR$ is a parallelogram.
$2$. Consider $PQCR$: Since $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = RC$. Also,$QR \parallel AB$ and $QR = \frac{1}{2} AB = PC$. Since $PQ \parallel RC$ and $QR \parallel PC$,$PQCR$ is a parallelogram.
$3$. Consider $PQRA$: Since $Q$ is the midpoint of $BC$ and $R$ is the midpoint of $AC$,$QR \parallel AB$ and $QR = \frac{1}{2} AB = AP$. Also,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC = AR$. Since $QR \parallel AP$ and $PQ \parallel AR$,$PQRA$ is a parallelogram.
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