The rectangle

Definition: a rectangle is a parallelogram with a right angle.

A rectangle is a parallelogram, so it has all the properties of a parallelogram:

• its opposite sides are parallel and equal;
• its opposite angles are equal, and two adjacent angles add up to $\displaystyle 180^{\circ}$;
• its diagonals bisect each other.

Theorem: all the angles of a rectangle are right.

In the rectangle $\displaystyle ABCD$, $\displaystyle {\angle A}={90^{\circ}}$.

• $\displaystyle {\angle B}={180^{\circ}-90^{\circ}}={90^{\circ}}$, since two adjacent angles of a parallelogram add up to $\displaystyle 180^{\circ}$;
• $\displaystyle {\angle C}={\angle A}$ and $\displaystyle {\angle D}={\angle B}$, since the opposite angles of a parallelogram are equal.

Theorem: the diagonals of a rectangle are equal.

• In the rectangle $\displaystyle ABCD$, $\displaystyle {AC}={BD}$.
• The diagonals also bisect each other. If they meet at the point $\displaystyle O$, the four halves are equal: $\displaystyle {AO}={BO}={CO}={DO}$.

Given: $\displaystyle ABCD$ is a rectangle.
Prove: $\displaystyle {AC}={BD}$.

The diagonals are sides of the triangles $\displaystyle DAB$ and $\displaystyle CBA$, so the proof below shows that these triangles are congruent:

In the rectangle $\displaystyle ABCD$, $\displaystyle {AB}={8}$ and $\displaystyle {BC}={6}$.
The diagonals meet at the point $\displaystyle O$.

• In triangle $\displaystyle ABC$ the angle $\displaystyle B$ is right. By Pythagoras’ theorem $\displaystyle {AC^2}={8^2+6^2}={100}$, so $\displaystyle {AC}={10}$;
• the diagonals are equal, so $\displaystyle {BD}={10}$ as well;
• the diagonals bisect each other, so $\displaystyle {AO}={BO}={CO}={DO}={5}$.

Two converse theorems prove that a quadrilateral is a rectangle:

Theorem: a parallelogram whose diagonals are equal is a rectangle.

Theorem: a quadrilateral with three right angles is a rectangle.

Given: $\displaystyle ABCD$ is a parallelogram, and $\displaystyle {AC}={BD}$.
Prove: $\displaystyle ABCD$ is a rectangle.

The proof below shows that the angle $\displaystyle DAB$ is right: the triangles $\displaystyle DAB$ and $\displaystyle CBA$ are congruent, so $\displaystyle {\angle DAB}={\angle CBA}$. These are two adjacent angles of a parallelogram, and they add up to $\displaystyle 180^{\circ}$.

In the quadrilateral $\displaystyle ABCD$, $\displaystyle {\angle A}={\angle B}={\angle C}={90^{\circ}}$.

• The angles $\displaystyle A$ and $\displaystyle B$ are co-interior angles between the lines $\displaystyle AD$ and $\displaystyle BC$. They add up to $\displaystyle 180^{\circ}$, so $\displaystyle AD\parallel BC$;
• in the same way, the angles $\displaystyle B$ and $\displaystyle C$ add up to $\displaystyle 180^{\circ}$, so $\displaystyle AB\parallel DC$;
• so $\displaystyle ABCD$ is a parallelogram with a right angle: a rectangle.

A quadrilateral is a rectangle if one of these holds:

• it is a parallelogram with a right angle (the definition);
• it is a parallelogram whose diagonals are equal;
• it has three right angles.

$\displaystyle ABCD$ is a rectangle, and its diagonals meet at the point $\displaystyle O$.
Work it out and give the reasons.

$\displaystyle {AC}={14}$.
Find $\displaystyle BO$.

$\displaystyle {\angle OAB}={35^{\circ}}$.
Find the angles $\displaystyle AOB$ and $\displaystyle OBC$.

$\displaystyle {AB}={12}$ and $\displaystyle {AD}={5}$.
Find $\displaystyle AC$ and the perimeter of triangle $\displaystyle AOB$.

$\displaystyle ABCD$ is a rectangle, and the point $\displaystyle M$ is the midpoint of the side $\displaystyle AB$.
Prove: $\displaystyle {MD}={MC}$.

$\displaystyle ABCD$ is a parallelogram, and its diagonals meet at the point $\displaystyle O$.
$\displaystyle {OA}={OB}$.
Prove: $\displaystyle ABCD$ is a rectangle.