Theorems About Parallelograms

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 $\overline{RS}.$

This theorem can be proven by using congruent triangles. Consider the parallelogram $PQRS$ and its diagonals $\overline{PR}$ and $\overline{QS}.$ Let $M$ be the point intersection of the diagonals.

Since $\overline{PQ}$ and $\overline{SR}$ are parallel, by the Alternate Interior Angles Theorem it can be stated that $\angle QPR\cong \angle SRP$ and that $\angle PQS\cong \angle RSQ.$ Furthermore, by the Parallelogram Opposite Sides Theorem it can be said that $\overline{PQ}\cong \overline{SR}.$

Here, two angles of $△PMQ$ and their included side are congruent to two angles of $△RMS$ and their included side. Therefore, by the Angle-Side-Angle Congruence Theorem $△PMQ$ and $△RMS$ are congruent triangles. Since corresponding parts of congruent triangles are congruent, $\overline{PM}$ is congruent to $\overline{RM}$ and $\overline{QM}$ is congruent to $\overline{SM}.$

By the definition of a segment bisector, both segments $\overline{PR}$ and $\overline{QS}$ are bisected at point $M.$ Therefore, it has been proven that the diagonals of a parallelogram bisect each other.

This proof will use similar triangles to prove the theorem.

Part I: $PQRS$ Is a Rectangle $\Rightarrow \overline{PR}\cong \overline{QS}$

Suppose $PQRS$ is a rectangle and $\overline{PR}$ and $\overline{QS}$ are its diagonals. By the Parallelogram Opposite Sides Theorem, the opposite sides of a parallelogram are congruent. Therefore, $\overline{RS}$ and $\overline{QP}$ are congruent. Additionally, by the Reflexive Property of Congruence, $\overline{SP},$ or $\overline{PS},$ is congruent to itself.

Since the angles of a rectangle are right angles, by the definition of congruent angles, $\angle RSP\cong \angle QPS$. Consequently, $△RSP$ and $△QPS$ have two pairs of congruent sides and congruent included angles. Therefore, by the Side-Angle-Side Congruence Theorem, the triangles are congruent. Because corresponding parts of congruent triangles are congruent, $\overline{PR}$ and $\overline{QS},$ which are the diagonals of $PQRS,$ are congruent.

Part II: $\overline{PR}\cong \overline{QS}\Rightarrow PQRS$ Is a Rectangle

Consider the parallelogram $PQRS$ and its diagonals $\overline{PR}$ and $\overline{QS}$ such that $\overline{PR}\cong \overline{QS}.$

By the Parallelogram Opposite Sides Theorem, $\overline{PQ}\cong \overline{SR}.$ Additionally, by the Reflexive Property of Congruence, $\overline{PS}$ is congruent to itself.

The sides of $△QPS$ are congruent to the sides of $△RSP.$ Therefore, by the Side-Side-Side Congruence Theorem, $△QPS\cong △RSP.$ Moreover, since corresponding parts of congruent triangles are congruent, $\angle QPS$ is congruent to $\angle RSP.$ Note that $\angle QPS$ and $\angle RSP$ are consecutive angles. By the Parallelogram Consecutive Angles Theorem, these angles are supplementary. With this information, it can be concluded that both $\angle QPS$ and $\angle RSP$ are right angles. Additionally, by the Parallelogram Opposite Angles Theorem, $\angle QPS\cong \angle SRQ$ and $\angle RSP\cong PQR.$ Because all of the angles are right angles, $PQRS$ is a rectangle.

This proof will use transformations to prove the theorem.

Part I: $PQRS$ Is a Rectangle $\Rightarrow \overline{PR}\cong \overline{QS}$

Consider the rectangle $PQRS$ and its diagonals $\overline{PR}$ and $\overline{QS}.$ Let $M$ be the point of intersection of the diagonals.

Let $A$ and $B$ be the midpoints of $\overline{PS}$ and $\overline{RQ}.$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

Note that $\overline{QA},$ $\overline{AR},$ $\overline{PB},$ and $\overline{BS}$ are congruent segments. Because congruent segments have the same length, the distance between $Q$ and $A$ equals the distance between $R$ and $A.$ Therefore, $Q$ is the image of $R$ after a reflection across $\overleftrightarrow{AB}.$ Similarly, $P$ is the image of $S$ after the same reflection. Since $M$ lies on $\overleftrightarrow{AB},$ a reflection across $\overleftrightarrow{AB}$ maps $M$ onto itself.

Reflection Across $\overleftrightarrow{AB}$
Preimage	Image
$R$	$Q$
$S$	$P$
$M$	$M$

The table shows that the images of the vertices of $△RSM$ are the vertices of $△QPM.$ Therefore, $△QPM$ is the image of $△RSM$ after a reflection across $\overleftrightarrow{AB}.$ Since a reflection is a rigid motion, this proves that the triangles are congruent. Because corresponding parts of congruent figures are congruent, $\overline{QM}\cong \overline{RM}$ and $\overline{PM}\cong \overline{SM}.$ Additionally, by the Parallelogram Diagonals Theorem, the diagonals of the rectangle bisect each other. Therefore, all four segments are congruent. Each diagonal of the parallelogram consists of the same two congruent segments. By the Segment Addition Postulate, the diagonals are congruent.

Part II: $\overline{PR}\cong \overline{QS}\Rightarrow PQRS$ Is a Rectangle

Consider the parallelogram $PQRS$ and its diagonals $\overline{PR}$ and $\overline{QS}$ such that $\overline{PR}\cong \overline{QS}.$ By the Parallelogram Diagonals Theorem, the diagonals of a rectangle bisect each other at $M.$

By the Parallelogram Opposite Sides Theorem, $\overline{PQ}\cong \overline{SR}$ and $\overline{QR}\cong \overline{PS}.$

Let $A$ and $B$ be the midpoints of $\overline{PS}$ and $\overline{RQ}.$ Then, a line through $M$ and the midpoints $A$ and $B$ can be drawn.

As shown before, $Q,$ $P,$ and $M$ are the respective images of $R,$ $S,$ and $M$ after a reflection across $\overleftrightarrow{AB}.$ Therefore, since $△QPM$ is the image of $△RSM$ after a reflection across $\overleftrightarrow{AB},$ the triangles are congruent. Let $C$ and $D$ be the midpoints of of $\overline{PQ}$ and $\overline{RS}.$ By following the same reasoning, $△PMS$ is the image of $△QMR$ after a reflection across $\overleftrightarrow{CD}.$ Therefore, the triangles are congruent. The parallelogram consists of four triangles in which the opposite triangles are congruent. Therefore, the corresponding angles of these triangles are congruent. Additionally, because all triangles are isosceles, the angles opposite congruent sides are congruent as well. Each angle of the parallelogram is the sum of the same two congruent angles. Therefore, all angles of the parallelogram are congruent. Moreover, by the Parallelogram Consecutive Angles Theorem, $\angle P$ and $\angle S$ are supplementary. With this information, it can be concluded that both angles are right triangles. Because all of the angles are congruent, the angles of the parallelogram are right triangles. Therefore, $PQRS$ is a rectangle.