Parallelogram Opposite Sides Theorem

This theorem can be proven by placing the parallelogram on a coordinate plane. For simplicity, vertex $P$ will be placed at the origin and vertex $S$ on the $x\text{-}$axis.

Since $P$ lies on the origin, its coordinates are $(0,0).$ Point $S$ is on the $x\text{-}$axis, meaning its $y\text{-}$coordinate is $0.$ Let $a$ be the $x\text{-}$coordinate of $S.$ Furthermore, let $b$ and $c$ be the coordinates of $Q.$ Note that both $P$ and $S$ lie on the $x\text{-}$axis. Therefore, $\overline{SP}$ is a horizontal segment. Since opposite sides of a parallelogram are parallel, $\overline{QR}$ is also a horizontal segment. This means that $Q$ and $R$ have the same $y\text{-}$coordinate. Let $x$ be the $x\text{-}$coordinate of $R.$ Next, the $x\text{-}$coordinate of $R$ will be determined. Since $\overline{PQ}$ and $\overline{SR}$ are parallel, they have the same slope. The slope of $\overline{PQ}$ can be found using the Slope Formula. The slope of $\overline{PQ}$ is $\frac{c}{b}.$ By following the same procedure, the slope of $\overline{SR}$ can be expressed in terms of $x.$

$m=\frac{y_2-y_1}{x_2-x_1}$
Side	Endpoints	Substitute	Simplify
$\overline{PQ}$	$P(0,0)$ and $Q(b,c)$	$m_{\overline{PQ}}=\frac{c-0}{b-0}$	$m_{\overline{PQ}}=\frac{c}{b}$
$\overline{SR}$	$S(a,0)$ and $R(x,c)$	$m_{\overline{SR}}=\frac{c-0}{x-a}$	$m_{\overline{SR}}=\frac{c}{x-a}$

As it has been previously stated, since $\overline{PQ}$ and $\overline{SR}$ are parallel, their slopes are equal. The above equation can be solved for $x.$ The $x\text{-}$coordinate of $R$ is $a+b.$ Finally, by using the Distance Formula, the length of each side of the parallelogram can be calculated. The length of $\overline{PQ}$ will be calculated first. By following the same procedure, all the side lengths can be calculated.

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Side	Endpoints	Substitute	Simplify
$\overline{PQ}$	$P(0,0)$ and $Q(b,c)$	$PQ=$ $\sqrt{(b-0{)}^2+(c-0{)}^2}$	$PQ=\sqrt{b^2+c^2}$
$\overline{QR}$	$Q(b,c)$ and $R(a+b,c)$	$QR=$ $\sqrt{(a+b-b{)}^2+(c-c{)}^2}$	$QR=a$
$\overline{SR}$	$S(a,0)$ and $R(a+b,c)$	$SR=$ $\sqrt{(a+b-a{)}^2+(c-0{)}^2}$	$SR=\sqrt{b^2+c^2}$
$\overline{PS}$	$P(0,0)$ and $S(a,0)$	$PS=$ $\sqrt{(a-0{)}^2+(0-0{)}^2}$	$PS=a$

By the Transitive Property of Equality, it can be said that $PQ=SR$ and that $QR=PS.$ By definition of congruent segments, it can be stated that the opposite sides of a parallelogram are congruent.

This theorem can also be proven by using congruent triangles. Consider the parallelogram $PQRS$ and its diagonal $\overline{PR}.$

It can be noted that two triangles are formed with $\overline{PR}$ as a common side. By the definition of a parallelogram, $\overline{PQ}$ and $\overline{SR}$ are parallel. Therefore, by the Alternate Interior Angles Theorem, it can be stated that $\angle QPR\cong \angle SRP$ and that $\angle QRP\cong \angle SPR.$ Furthermore, by the Reflexive Property of Congruence, $\overline{PR}$ is congruent to itself. Consequently, $△PQR$ and $△RSP$ have two pairs of congruent angles and an included congruent side. Therefore, by the Angle-Side-Angle Congruence Theorem, $△PQR$ and $△RSP$ are congruent triangles. Since corresponding parts of congruent figures are congruent, $\overline{PS}$ is congruent to $\overline{QR}$ and $\overline{PQ}$ is congruent to $\overline{RS}.$