Rectangle: definition, properties and diagonals

A rectangle is defined as a parallelogram in which every interior angle is a right angle. Because it is built on top of "parallelogram," a rectangle automatically inherits everything true of parallelograms in general, and then adds its own signature property on top.

Inherited from being a parallelogram. Opposite sides are equal and parallel, and the diagonals bisect each other. If $ABCD$ is a rectangle, then $AB = CD$, $BC = AD$, $AB \parallel CD$, $BC \parallel AD$, and if the diagonals $AC$ and $BD$ meet at $O$, then $O$ is the midpoint of both diagonals, so $AO = OC$ and $BO = OD$.

Unique to the rectangle. All four angles are exactly $90°$: $\angle A = \angle B = \angle C = \angle D = 90°$. This right-angle condition, combined with the parallelogram's own symmetry, forces the two diagonals to come out equal in length, not just bisecting — a property a slanted (non-rectangular) parallelogram does not have. Putting the bisection and the equal-length facts together gives one compact statement about the diagonals, useful for solving almost any rectangle-diagonal question:

where: $AC$ and $BD$ are the two diagonals, $O$ is their point of intersection, and $AO$, $OD$ are the half-diagonal segments from $O$ to a vertex — since $O$ is the midpoint of each diagonal, doubling any one half-segment reproduces the full diagonal length. It is worth being precise about what this property does not say. The diagonals of a rectangle bisect each other, but in a non-square rectangle they do not bisect the vertex angles (a diagonal does not split a $90°$ corner into two $45°$ halves unless the rectangle is a square), and they are not perpendicular to each other — perpendicular diagonals is a rhombus property, and only a square, being both a rectangle and a rhombus, has diagonals that are equal, bisecting, and perpendicular all at once.

The two properties feeding the boxed relation are not independent tricks — one comes from "rectangle is a parallelogram" and the other from "rectangle also has right angles," and together they pin down every diagonal length or segment once a single piece of information is known.

Step 1. Bisection comes purely from the parallelogram structure: in any parallelogram, opposite sides being equal and parallel forces the diagonals to cross at their common midpoint, so $AO = OC$ and $BO = OD$ regardless of whether any angle is a right angle.

Step 2. Equal diagonal length comes specifically from the right angles: in triangle $ABC$ of rectangle $ABCD$, angle $B$ is $90°$, and comparing triangle $ABC$ with triangle $DCB$ (which share side $BC$, have $AB = DC$, and both contain a right angle at $B$ and $C$ respectively) shows the two diagonals $AC$ and $BD$ must be equal — a fact that fails the moment the angle is not $90°$, which is exactly why a slanted parallelogram's diagonals are generally unequal.

Step 3. Combining the two gives the working formula: because $O$ bisects both diagonals (Step 1) and the diagonals are equal to begin with (Step 2), all four half-segments $AO$, $OC$, $BO$, $OD$ are equal to each other, and doubling any single one reproduces both full diagonals: $AC = BD = 2\times AO = 2\times OD$.

Problem. $PQRS$ is a rectangle with diagonals meeting at point $O$. If $PO = 6.5\text{ cm}$, find the lengths of both diagonals $PR$ and $QS$.

Step 1. Recall the two relevant properties: in a rectangle, the diagonals bisect each other, and the diagonals are equal in length.

Step 2. Use bisection to find the full length of one diagonal. Since $O$ is the midpoint of diagonal $PR$, $PO = OR = 6.5\text{ cm}$, so

Step 3. Use the equal-diagonals property to find the other diagonal. Since the diagonals of a rectangle are equal, $QS = PR = 13\text{ cm}$.

Final answer: $PR = 13\text{ cm}$ and $QS = 13\text{ cm}$.

Competitive stretch

Problem. In rectangle $ABCD$, $\angle AOB = 70°$, where $O$ is the intersection of the diagonals. Find $\angle BOC$, $\angle COD$, and $\angle AOD$.

Step 1. Around point $O$, the four angles $\angle AOB$, $\angle BOC$, $\angle COD$, $\angle DOA$ always add up to $360°$.

Step 2. $\angle AOB$ and $\angle COD$ are vertically opposite (the diagonals are straight lines crossing at $O$), so $\angle COD = \angle AOB = 70°$; similarly $\angle BOC = \angle AOD$.

Step 3. $\angle AOB$ and $\angle BOC$ lie on the straight diagonal $AC$, so they are supplementary: $\angle AOB + \angle BOC = 180°$, giving $\angle BOC = 180° - 70° = 110°$, and hence $\angle AOD = 110°$ too.

Final answer: $\angle BOC = 110°$, $\angle COD = 70°$, $\angle AOD = 110°$. This reasoning uses only vertical and linear-pair angle facts true of any two crossing diagonals; it does not rely on the rectangle's equal-diagonal property, since a rectangle's diagonals need not cross at right angles unless it is also a square.

• Assuming rectangle diagonals are perpendicular. They are not, in general — perpendicular diagonals is a rhombus property; only a square has diagonals that are both equal and perpendicular.
• Assuming rectangle diagonals bisect the vertex angles. In a non-square rectangle a diagonal does not split a $90°$ corner into two equal $45°$ parts.
• Forgetting "equal" is extra, beyond parallelogram properties. "Diagonals bisect each other" is true for every parallelogram; the diagonals also being equal is what makes a rectangle special, and it is easy to state only the first fact.
• Mixing up a rectangle with a square. Every square is a rectangle, but not every rectangle is a square — a rectangle only needs four right angles, not four equal sides.

Ask the tutor: Want to try a mixed diagonal-length and diagonal-angle problem, or see exactly why the diagonals-equal property breaks down for a general (non-rectangular) parallelogram?