Understanding Quadrilaterals

Look around any room and quadrilaterals are everywhere: the rectangular door, the square tile, the kite-shaped paper toy, the slanted trapezium of a laptop screen tilted back. All of these are four-sided closed figures, yet they behave very differently — a door's diagonals are equal, a kite's diagonals cross at a right angle, a general four-sided field has diagonals that do nothing special at all. This chapter builds the vocabulary and the logic to tell these shapes apart with proof, not just by looking. We start from the widest idea, the polygon — any simple closed figure made only of line segments — and narrow down in careful steps: convex versus concave, regular versus irregular, and then the special four-sided family of trapeziums, kites and parallelograms. The single most useful fact in the whole chapter is almost embarrassingly simple: if you walk once around the boundary of any convex polygon, turning a little at each corner, you turn through exactly one full circle. That one observation — the exterior angle sum is $360^\circ$ — unlocks formulas for the interior angles of any polygon, regular or not, with any number of sides. The second half of the chapter turns to parallelograms, the four-sided shapes whose opposite sides never meet. Using nothing more than triangle congruence rules already learned, we prove — not merely observe — that a parallelogram's opposite sides are equal, its opposite angles are equal, its adjacent angles are supplementary, and its diagonals bisect each other. From there, the rhombus, the rectangle and the square fall out as parallelograms with one extra condition each, each inheriting every earlier property and adding its own. By the end you will be able to name any quadrilateral from its diagonals alone, find missing angles and side lengths using algebra, and explain in one sentence why a square is simultaneously a rhombus, a rectangle and a parallelogram.

Key takeaways

A polygon is a simple closed figure made of line segments; it is convex if no diagonal or interior segment ever steps outside it, and concave otherwise.
A regular polygon is both equilateral (equal sides) and equiangular (equal angles); an irregular polygon fails at least one of these.
The sum of the exterior angles of any convex polygon, however many sides it has, is always $360^\circ$ .
The sum of the interior angles of a convex polygon with $n$ sides is $(n-2) \times 180^\circ$ , derived directly from the exterior angle sum.
A trapezium has exactly one pair of parallel sides; a kite has two distinct pairs of equal consecutive sides and diagonals that meet at right angles.
A parallelogram has both pairs of opposite sides parallel, which forces opposite sides equal, opposite angles equal, adjacent angles supplementary, and diagonals that bisect each other.
A rhombus is a parallelogram with all sides equal, so its diagonals are also perpendicular bisectors of each other.
A rectangle is a parallelogram with one right angle (hence all four are right angles), so its diagonals are equal in length as well as bisecting.
A square is a rectangle with all sides equal — equivalently a rhombus with all angles equal — so it carries every property above: equal sides, right angles, and diagonals that are equal, perpendicular, and bisecting.
Key idea: every special quadrilateral in this chapter is a parallelogram (or kite) with one extra condition added, and each extra condition can be proved, using triangle congruence, to force a matching extra property in the diagonals.

The core insight

A convex polygon's exterior angles always total $360^\circ$ , and a quadrilateral earns the name parallelogram, rhombus, rectangle or square by satisfying one more condition than the last, each condition provable by triangle congruence and each adding a predictable new property to its diagonals.

What this chapter covers

This article follows the chapter's own sequence. First, polygons in general — what makes a curve a polygon, and the convex/concave and regular/irregular splits, together with the exterior-angle-sum result and the interior-angle formula it produces. Next, the basic special quadrilaterals that are not parallelograms: the trapezium (one pair of parallel sides) and the kite (two pairs of equal consecutive sides). Then the heart of the chapter, the parallelogram, built up property by property — equal opposite sides, equal opposite angles, supplementary adjacent angles, and diagonals that bisect each other — each one proved using congruent triangles rather than just observed. Finally, the special parallelograms — rhombus, rectangle and square — each defined as a parallelogram with one extra condition, with the resulting extra diagonal property proved in each case. The chapter closes by tying every shape together into a single hierarchy, which is exactly what the worked examples and the cheatsheet at the end are built to test.

The core ideas

Polygons, convex and concave, regular and irregular

A polygon is a simple closed curve made up only of line segments. It is convex if every diagonal lies entirely inside the figure and no interior angle is reflex (greater than $180^\circ$ ); it is concave if some corner caves in, pushing part of a diagonal outside. This chapter works only with convex polygons. A polygon is regular when it is both equilateral (equal sides) and equiangular (equal angles) — a square qualifies, but a rectangle does not, since its angles are equal but its sides are not. An irregular polygon fails at least one of these.

The exterior angle sum of a polygon

Walk along the sides of any convex polygon, turning at each corner to face the next side. By the time you return to the start you have turned through exactly one complete revolution, whatever the number of sides.

\angle 1 + \angle 2 + \angle 3 + \cdots + \angle n = 360^\circ. \qquad (1)

where: $\angle 1, \ldots, \angle n$ are the exterior angles at the $n$ vertices, each formed between a side and the extension of the next side. This holds for a triangle, a hexagon, or a hundred-sided polygon alike.

Figure

A convex pentagon with an exterior angle marked at each vertex beside a circle divided into five matching wedges that together make one full turn of 360 degrees

The interior angle sum of a polygon

At every vertex, the interior angle and the exterior angle sit on a straight line and so add to $180^\circ$ . Summing this over all $n$ vertices and subtracting the exterior angle total of $360^\circ$ gives the interior angle sum directly — worked out in full as the first derivation below.

S_{\text{interior}} = (n-2) \times 180^\circ. \qquad (2)

where: $S_{\text{interior}}$ is the sum of all $n$ interior angles of a convex polygon with $n$ sides. For a quadrilateral, $n = 4$ , so $S_{\text{interior}} = (4-2)\times 180^\circ = 360^\circ$ — the four angles of any quadrilateral, however irregular, always add to $360^\circ$ . For a regular polygon, divide by $n$ to get each individual angle:

\text{each exterior angle} = \frac{360^\circ}{n}, \qquad \text{each interior angle} = 180^\circ - \frac{360^\circ}{n}. \qquad (3)

where: $n$ is the number of sides of the regular polygon; both formulas assume all $n$ angles are equal, which is only guaranteed when the polygon is regular.

Trapezium and kite

A trapezium is a quadrilateral with exactly one pair of parallel sides. If the two non-parallel sides are also equal in length, it is an isosceles trapezium. A trapezium is not a parallelogram, since only one pair of sides is parallel. A kite is a quadrilateral with two distinct pairs of consecutive equal sides — for kite $ABCD$ with $AB = AD$ and $CB = CD$ , the equal pairs meet at a vertex rather than sitting opposite each other, unlike a parallelogram. A kite's diagonals meet at right angles, and only one diagonal is bisected by the other; the two angles between the unequal sides ( $\angle B$ and $\angle D$ ) are equal, but the other two angles are not, in general.

The parallelogram and its four properties

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. This single condition forces four separate, provable consequences: opposite sides equal, opposite angles equal, adjacent angles supplementary (parallel sides with a transversal make co-interior angles add to $180^\circ$ ), and diagonals that bisect each other at their crossing point. The full proof chain is worked through step by step in the second derivation below.

\angle A + \angle B = 180^\circ, \qquad \angle B + \angle C = 180^\circ, \qquad \text{and so on around the figure.} \qquad (4)

where: $\angle A, \angle B, \angle C, \angle D$ are the four interior angles of parallelogram $ABCD$ taken in order, so each pair sharing a side (adjacent angles) is supplementary, while each pair of opposite angles, $\angle A, \angle C$ and $\angle B, \angle D$ , is equal.

Figure

A tree diagram showing the quadrilateral family with the quadrilateral box at the top branching to trapezium, kite and parallelogram, and parallelogram branching further into rhombus and rectangle which both connect down into a shared square at the bottom

Special parallelograms — rhombus, rectangle, square

A rhombus is a parallelogram with all four sides equal. It keeps every parallelogram property, and the equal sides add one more: its diagonals are perpendicular bisectors of each other, meeting at exactly $90^\circ$ . A rectangle is a parallelogram with one right angle — and once one angle is $90^\circ$ , the supplementary-adjacent-angle property forces every angle to be $90^\circ$ . It adds one more property: its diagonals are equal in length, besides bisecting each other. A square is a rectangle with all sides equal — equivalently, a rhombus with all angles equal. It is where every condition meets at once, so it inherits everything: all sides equal, all angles right angles, and diagonals that are simultaneously equal, perpendicular, and bisecting.

OA = OC, \quad OB = OD \ \ (\text{always, parallelogram}); \qquad AC = BD \ \ (\text{rectangle, square}); \qquad AC \perp BD \ \ (\text{rhombus, square}). \qquad (5)

where: $O$ is the point where the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ cross; the first pair of equalities holds in every parallelogram, the equal-length condition is the extra fact true only for rectangles and squares, and the perpendicularity is the extra fact true only for rhombuses and squares.

Angle sum from turning — the exterior-to-interior bridge

Derivation AInterior angle sum of a polygon from the exterior angle sum

The goal is to derive the general formula $S_{\text{interior}} = (n-2)\times 180^\circ$ for the interior angle sum of any convex polygon with $n$ sides, starting only from the turning argument's fact that exterior angles always add to $360^\circ$ , plus the observation that an interior angle and its exterior angle at the same vertex add to a straight angle of $180^\circ$ . No geometry beyond linear pairs and one algebraic rearrangement is needed.

Step 1. Set up one equation per vertex. At each vertex the interior angle and the exterior angle are supplementary, since the exterior angle is formed by extending one side, and the two together make a straight line. So at vertex $k$ , interior angle $I_k$ and exterior angle $E_k$ satisfy $I_k + E_k = 180^\circ$ .

Step 2. Add this equation over all $n$ vertices. Summing gives $(I_1 + \cdots + I_n) + (E_1 + \cdots + E_n) = n \times 180^\circ$ .

Step 3. Substitute the known exterior angle sum. The turning argument gives $E_1 + \cdots + E_n = 360^\circ$ for any convex polygon. Substituting, $S_{\text{interior}} + 360^\circ = n \times 180^\circ$ , where $S_{\text{interior}}$ is the interior angle sum we want.

Step 4. Solve for it. Rearranging, $S_{\text{interior}} = n \times 180^\circ - 360^\circ = (n-2) \times 180^\circ$ , since $360^\circ = 2 \times 180^\circ$ . No assumption of regularity was used — only the linear-pair relation at each vertex and the constant exterior angle total.

Step 5. Check against small cases. For a triangle, $n=3$ gives $180^\circ$ , the familiar angle sum of a triangle. For a quadrilateral, $n=4$ gives $360^\circ$ ; for a hexagon, $n=6$ gives $720^\circ$ — both match splitting the polygon into triangles from one vertex.

Go deeper · Derivation B · 1 minParallelogram properties from one diagonal and triangle congruence

Derivation BParallelogram properties from one diagonal and triangle congruence

The goal is to prove, not merely observe, that in any parallelogram $ABCD$ the opposite sides are equal and the diagonals bisect each other, starting only from the defining condition — both pairs of opposite sides parallel — and the ASA and SSS congruence rules. This is the logical backbone used later to prove the extra rhombus, rectangle and square properties by the same method.

Step 1. Draw one diagonal and name the angles it creates. In parallelogram $ABCD$ , draw diagonal $AC$ . Since $AB \parallel DC$ with transversal $AC$ , alternate angles give $\angle BAC = \angle DCA$ . Since $AD \parallel BC$ with the same transversal, alternate angles also give $\angle DAC = \angle BCA$ .

Step 2. Apply ASA to the two triangles the diagonal creates. Triangles $ABC$ and $CDA$ share side $AC$ , and Step 1 gives a pair of equal angles at $A$ and at $C$ on either side of that shared side. By ASA (angle, included side, angle), $\triangle ABC \cong \triangle CDA$ .

Step 3. Read off equal sides. Because $\triangle ABC \cong \triangle CDA$ , corresponding sides are equal: $AB = CD$ and $BC = DA$ — opposite sides of a parallelogram are equal, proved from the parallel-sides definition alone.

Step 4. Draw the second diagonal and isolate a smaller pair of triangles. Draw diagonal $BD$ , meeting $AC$ at $O$ . In triangles $AOB$ and $COD$ : transversal $BD$ gives $\angle ABO = \angle CDO$ , transversal $AC$ gives $\angle BAO = \angle DCO$ , and Step 3 gives the included side $AB = CD$ .

Step 5. Apply ASA again and conclude. With $\angle BAO = \angle DCO$ , $AB = CD$ , and $\angle ABO = \angle CDO$ , $\triangle AOB \cong \triangle COD$ by ASA. So $AO = CO$ and $BO = DO$ — the diagonals bisect each other at $O$ . Every later property (equal diagonals for a rectangle, perpendicular diagonals for a rhombus) is proved by one more pair of congruent triangles built on this same base result.

Worked examples

Board level - exterior angles of a pentagon

The exterior angles of a pentagon, taken in order, measure $70^\circ, 60^\circ, 65^\circ, 80^\circ$ and $x$ . Find $x$ .

Step 1. Use the exterior angle sum property: the exterior angles of any convex polygon add to $360^\circ$ . So $70^\circ + 60^\circ + 65^\circ + 80^\circ + x = 360^\circ$ .

Step 2. Add the known angles: $70 + 60 + 65 + 80 = 275$ , giving $275^\circ + x = 360^\circ$ .

Step 3. Solve: $x = 360^\circ - 275^\circ = 85^\circ$ . Answer: $x = 85^\circ$ .

Board level - a parallelogram diagonal problem with algebra

$HELP$ is a parallelogram whose diagonals meet at $O$ . Given that $OE = 4$ cm and that the full length $HL$ is $5$ cm more than the full length $PE$ , find $OH$ .

Step 1. Diagonals of a parallelogram bisect each other, so $O$ is the midpoint of diagonal $PE$ : $OP = OE = 4$ cm.

Step 2. Find the full diagonal $PE = OP + OE = 4 + 4 = 8$ cm.

Step 3. Use the given relation: $HL$ is $5$ cm more than $PE$ , so $HL = 8 + 5 = 13$ cm.

Step 4. $OH$ is half of $HL$ , again by bisection: $OH = \tfrac{1}{2}\times 13 = 6.5$ cm. Answer: $OH = 6.5$ cm.

Higher-order thinking (HOTS / olympiad-style) - proving a quadrilateral must be a rectangle

In quadrilateral $PQRS$ , the diagonals $PR$ and $QS$ bisect each other at a point $O$ , and are also equal in length. Prove that $PQRS$ must be a rectangle.

Step 1. Show $PQRS$ is a parallelogram first. Since diagonals bisect each other, $OP = OR$ and $OQ = OS$ ; in triangles $POQ$ and $ROS$ the included angles $\angle POQ = \angle ROS$ (vertically opposite). By SAS, $\triangle POQ \cong \triangle ROS$ , so $PQ = RS$ and $\angle OPQ = \angle ORS$ — alternate angles cut by transversal $PR$ — so $PQ \parallel RS$ . One pair of opposite sides both equal and parallel makes $PQRS$ a parallelogram.

Step 2. Bring in the equal-diagonals condition and compare a fresh pair of triangles: $PQR$ and $QPS$ . Given $PR = QS$ , plus $PQ = PQ$ (shared) and $QR = PS$ (opposite sides of the now-known parallelogram), SSS gives $\triangle PQR \cong \triangle QPS$ .

Step 3. Corresponding angles of congruent triangles are equal, so $\angle PQR = \angle QPS$ . But these are also adjacent parallelogram angles at side $PQ$ , so they are supplementary: $\angle PQR + \angle QPS = 180^\circ$ . Combining the two, $2\angle PQR = 180^\circ$ , so $\angle PQR = 90^\circ$ .

Step 4. A parallelogram with one right angle has all four angles right angles (adjacent angles supplementary, opposite angles equal), which is exactly the definition of a rectangle. Answer: $PQRS$ is a rectangle — bisecting diagonals make it a parallelogram, and equal diagonals then force every angle to be $90^\circ$ . Timing note: allow four to five minutes. This is a converse-style proof — you deduce the shape from diagonal properties, not the other way round. The trap is skipping Step 1: without first establishing $PQRS$ is a parallelogram, "opposite sides equal" and "adjacent angles supplementary" cannot legally be used in Steps 2 and 3.

Common traps

Treating a trapezium as a parallelogram. Only one pair of sides is parallel; do not apply "opposite sides equal" to it.
Forgetting convexity when adding exterior angles. The $360^\circ$ rule is stated for convex polygons; check the figure is convex first.
Mixing up interior and exterior angle formulas. Interior sum is $(n-2)\times 180^\circ$ ; each exterior angle of a regular polygon is $360^\circ/n$ — confusing the two costs a whole factor.
Assuming a kite's diagonals bisect each other. Only one diagonal of a kite is bisected by the other, unlike a parallelogram.
Using "diagonals bisect each other" as proof of "parallelogram" without checking. That is a consequence of being a parallelogram; to prove the shape from its diagonals you must argue it explicitly, as in the third worked example.
Sloppy congruence statements. Writing $\triangle ABC \cong \triangle CDA$ with mismatched vertex order breaks the correspondence between equal sides and angles — always match vertices in the same order on both sides of $\cong$ .

Common trapThe single biggest misconception is thinking every quadrilateral with some equal sides or some equal angles must automatically be a parallelogram, rhombus, rectangle or square

Being a parallelogram is not free from a coincidence of equal measurements — it comes only from the definition (both pairs of opposite sides parallel) or a logically equivalent proven condition, such as "one pair of opposite sides both equal and parallel" or "diagonals bisecting each other." A quadrilateral can have $AB = CD$ without $AB \parallel CD$ , and $\angle A = \angle C$ without $\angle B = \angle D$ . Every step from "some measurements match" to "this shape's name is X" must be justified by a definition or proved property — exactly the discipline the third worked example used to argue equal, bisecting diagonals into a full rectangle proof rather than assuming it.

You should be able to…

State the definition of a polygon, and classify a figure as convex or concave.
Explain what makes a polygon regular, and name the regular polygons with 3, 4, 5, 6, 8 and 20 sides.
Use the exterior angle sum ( $360^\circ$ ) to find a missing exterior angle or the number of sides of a regular polygon.
Use the interior angle sum formula $(n-2)\times 180^\circ$ to find a missing interior angle or the number of sides.
Distinguish a trapezium, kite and parallelogram by which sides are parallel or marked equal.
State and use all four parallelogram properties: opposite sides equal, opposite angles equal, adjacent angles supplementary, diagonals bisect each other.
Prove, using ASA or SSS congruence, that opposite sides of a parallelogram are equal and its diagonals bisect each other.
Explain the one extra condition each of rhombus, rectangle and square adds, and the matching extra diagonal property.
Solve algebraic problems using parallelogram or rhombus diagonal properties to find an unknown length or angle.
Argue a converse-style proof, such as showing bisecting and equal diagonals force a quadrilateral to be a rectangle.

Ask the tutor: If you are stuck on telling a parallelogram, rhombus, rectangle and square apart from their diagonals alone, ask the tutor to walk through the family tree with your own example figure.

Is every rhombus a square, and is every square a rhombus?

Not every rhombus is a square, but every square is a rhombus. A rhombus only requires all four sides equal; its angles can be anything, provided opposite angles match and adjacent angles are supplementary, so a "squashed" rhombus with angles $60^\circ, 120^\circ, 60^\circ, 120^\circ$ is valid and clearly not a square. A square, though, is a rectangle with all sides equal, so its angles are already forced to $90^\circ$ by the rectangle part of the definition — it automatically satisfies the rhombus condition too, plus the right angles. Square is the special case where rhombus and rectangle overlap.

Why do the diagonals of a parallelogram bisect each other but are not usually equal?

Bisection comes purely from the parallel-sides definition, proved by matching triangle $AOB$ with triangle $COD$ using ASA congruence in the second derivation above — no extra condition beyond "opposite sides parallel" is needed. Equal diagonal length is a stronger, separate fact that only follows once the parallelogram has a right angle, that is, once it is a rectangle. A "slanted" parallelogram, like a tilted picture frame, can have one diagonal much longer than the other while both are still cut exactly in half — bisection and equal length are independent properties, and only a rectangle guarantees both.

How can I quickly tell a kite from a rhombus in a figure?

Look at which sides are marked equal. In a rhombus, all four sides are equal, so the tick marks go all the way around. In a kite, only two pairs of consecutive sides are equal, so you see one tick mark on two sides meeting at one vertex, and a different (double) tick mark on the other two sides meeting at the opposite vertex — the two pairs are generally different lengths from each other. A rhombus is always a parallelogram; a kite is a parallelogram only in the special case where it also happens to be a rhombus, which is not true for a typical kite.

Why does the formula (n minus 2) times 180 degrees work for any polygon, not just regular ones?

The derivation used only two facts with nothing to do with regularity: an interior angle and its exterior angle at the same vertex always add to $180^\circ$ , and the exterior angles of any convex polygon add to $360^\circ$ from the "one full turn while walking around" argument. Neither fact assumes the sides or angles are equal — they hold for a regular hexagon or a wildly irregular seven-sided field boundary alike. That is why the formula works for every convex polygon with $n$ sides; only dividing by $n$ to get "each angle" requires the polygon to be regular.