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Convex quadrilaterals - parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides. Equivalent conditions are that opposite sides are of equal length; that opposite angles are equal; or that the diagonals bisect each other. Parallelograms also include the square, rectangle, rhombus and rhomboid.

• Rhombus or rhomb: all four sides are of equal length. Equivalent conditions are that opposite sides are parallel and opposite angles are equal, or that the diagonals perpendicularly bisect each other. An informal description is "a pushed-over square" (including a square).
• Rhomboid: a parallelogram in which adjacent sides are of unequal lengths and angles are oblique (not right angles). Informally: "a pushed-over rectangle with no right angles."
• Rectangle: all four angles are right angles. An equivalent condition is that the diagonals bisect each other and are equal in length. Informally: "a box or oblong" (including a square).
• Square (regular quadrilateral): all four sides are of equal length (equilateral), and all four angles are right angles. An equivalent condition is that opposite sides are parallel (a square is a parallelogram), that the diagonals perpendicularly bisect each other, and are of equal length. A quadrilateral is a square if and only if it is both a rhombus and a rectangle.
• Oblong: a term sometimes used to denote a rectangle which has unequal adjacent sides (i.e. a rectangle that is not a square).

A shape that is both a rhombus (four equal sides) and a rectangle (four equal angles) is a square (four equal sides and four equal angles).

Square → Rhombus → Parallelogram (opposite sides are parallel) → Quadrilateral (four-sided polygon)

Convex quadrilaterals - other

• Kite: two pairs of adjacent sides are of equal length. This implies that one diagonal divides the kite into congruent triangles, and so the angles between the two pairs of equal sides are equal in measure. It also implies that the diagonals are perpendicular. (It is common, especially in the discussions on plane tessellations, to refer to the concave quadrilateral with these properties as a dart or arrowhead, with term kite being restricted to the convex shape.)

• Trapezium (British English) or trapezoid (NAm.): one pair of opposite sides are parallel.
• Isosceles trapezium (Brit.) or isosceles trapezoid (NAm.): one pair of opposite sides are parallel and the base angles are equal in measure. This implies that the other two sides are of equal length, and that the diagonals are of equal length. An alternative definition is: "a quadrilateral with an axis of symmetry bisecting one pair of opposite sides".
• Trapezium (NAm.): no sides are parallel. (In British English this would be called an irregular quadrilateral, and was once called a trapezoid.)
• Cyclic quadrilateral: the four vertices lie on a circumscribed circle. A quadrilateral is cyclic if and only if opposite angles sum to 180°.
• Tangential quadrilateral: the four edges are tangential to an inscribed circle. Another term for a tangential polygon is inscriptible.
• Bicentric quadrilateral: both cyclic and tangential.

Area of a convex quadrilateral

There are various general formulas for the area of a convex quadrilateral.

The area of a quadrilateral ABCD can be calculated using vectors. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then

which is the magnitude of the cross product of vectors AC and BD. In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as:

and is equal to

,

where the lengths of the diagonals are p and q and the angle between them is θ.[2]

Bretschneider's formula[3] expresses the area in terms of the sides and angles:

,

where the sides in sequence are a,b,c,d, s is the semiperimeter, and γ and λ are any two opposite angles. This reduces to Brahmagupta's formula for the area of a cyclic quadrilateral when γ + λ = 180°.

Another area formula in terms of the sides and angles, with γ being between sides b and c and λ being between sides a and d, is

.

Next,[4] the following formula expresses the area in terms of the sides and diagonals:

,

where p and q are the diagonals. Again, this reduces to Brahmagupta's formula in the cyclic quadrilateral case, since then pq = ac + bd.

Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, so long as this angle is not 90°:[5]

.

In the case of a parallelogram, the latter formula becomes .

The area of a rhombus is given by for diagonals p,q.

The area of a parallelogram with sides a and b and angles C and 180°-C is .

The area of a cyclic quadrilateral with successive sides a, b, c, d and angle γ between sides b and c is

.

More quadrilaterals

• A geometric chevron (dart or arrowhead) is a concave quadrilateral with bilateral symmetry like a kite, but one interior angle is reflex.

• A self-intersecting quadrilateral is called variously a cross-quadrilateral, crossed quadrilateral, butterfly quadrilateral or bow-tie quadrilateral.

• A non-planar quadrilateral is called a skew quadrilateral.

Miscellaneous facts about quadrilaterals

• The diagonals of a rhombus bisect the angles.

• Let ABCD be a trapezoid (in the U.S. sense of having two parallel sides) with vertices A, B, C, and D in sequence and with parallel sides AB and DC. Let E be the intersection of the diagonals, and let F be on side DA and G be on side BC such that FEG is parallel to AB and CD. Then FG is the harmonic mean of AB and DC:

.

• A parallelogram with equal diagonals is a rectangle.

• A cyclic quadrilateral with successive sides a, b, c, d and diagonals p, q has pq=ac+bd.

• A cyclic quadrilateral with successive vertices A, B, C, D and successive sides a=AB, b=BC, c=CD, and d=DA and with diagonals p=AC and q=BD has:

, ,

and

.

• A cyclic quadrilateral with successive sides a, b, c, d and semiperimeter s has circumradius (the radius of the circumscribing circle) given by[6]

.

• A parallelogram with diagonals p, q and successive sides a, b, c, and d with d=b and c=a has

p2 + q2 = a2 + b2 + c2 + d2.

• For any point P in the interior of a rectangle with successive vertices A, B, C, D, we have

(AP)2 + (CP)2 = (BP)2 + (DP)2.

• Any line through the midpoint of a parallelogram bisects the area.

• The midpoints of the sides of a quadrilateral are the vertices of a parallelogram. The area of this inner parallelogram equals one-half the area of the outer quadrilateral. The perimeter of the inner parallelogram equals the sum of the diagonals of the outer quadrilateral.

• An orthodiagonal quadrilateral (one with perpendicular diagonals) with sides a, b, c, d in sequence has a2 + c2 = b2 + d2.

Taxonomy

A taxonomy of quadrilaterals is illustrated by the following graph. Lower forms are special cases of higher forms. Note that "trapezium" here is referring to the British definition (the North American equivalent is a trapezoid), and "kite" excludes the concave kite (arrowhead or dart). Inclusive definitions are used throughout.

References

What a great problem this is! I shared it with my department and they began to realize that it was challenging. One teacher tried to construct a counterexample using GeoSketchPad but that didn’t work out. Most believed it didn’t have to be a paralleogram because of the SSA case but demonstrating an example is much harder. Here was my approach:
I drew the diagonal connecting the non-congruent angles and proved using the Law of Sines that two of the ‘alternate interior’ angles formed had to be either equal (in which case it would be a parallelogram) or supplementary as in Darmok’s example (100 and 80!). However, I did not have Darmok’s persistence to try to draw the supplementary case by choosing a concrete example. I loved his construction. This morning I took his idea and carefully constructed a possible diagram starting with a side of 12 opposite the 100 degree angle and worked out all of the other sides using trig. I took my ruler and protractor, built it and shared it with everyone but i can’t take credit for it! Now can you name 5 of your students who would get this excited about this problem! I plan on cross-referencing all of this in my blog, MathNotations. Thanks jd2718 for challenging us!
Dave Marain