If one takes the middle number, 5, and multiplies it by 3 ( the base number of the magic square of three ), the result is 15, which is also the constant sum of all the rows, columns, and two main diagonals.

If M and N are the midpoints of the diagonals AC and BD, then

If M and N are the midpoints of the diagonals, then

If the trapezoid is divided into four triangles by its diagonals AC and BD ( as shown on the right ), intersecting at O, then the area of is equal to that of, and the product of the areas of and is equal to that of and.

If the sums of numbers on a magic cube's broken space diagonals also equal the cube's magic number, the cube is called a pandiagonal cube.

If rectangles are included in the class of trapezoids then one may concisely define an isosceles trapezoid as " a cyclic quadrilateral with equal diagonals " or as " a cyclic quadrilateral with a pair of parallel sides.

If a split-complex number z does not lie on one of the diagonals, then z has a polar decomposition.

If sides of Cyclic Quadrilateral are a, b, c, d and its diagonals are x and y while

: If a quadrilateral is inscribable in a circle then the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of opposite sides.

If now diameter AF is drawn bisecting DC so that DF and CF are sides c of an inscribed decagon, Ptolemy's Theorem can again be applied – this time to cyclic quadrilateral ADFC with diameter d as one of its diagonals:

From the criterion it also follows that Carmichael numbers are cyclic .< ref > Proof sketch: If is square-free but not cyclic, for two prime factors and of.

If negation is cyclic and "∨" is a " max operator ", then the law can be expressed in the object language by ( P ∨ ~ P ∨ ~~ P ∨ ... ∨ ~...~ P ), where "~...~" represents n − 1 negation signs and "∨ ... ∨" n − 1 disjunction signs.

If the computational Diffie – Hellman assumption holds in the underlying cyclic group, then the encryption function is one-way.

* If four points A, B, C, D are given that form a cyclic quadrilateral, then the nine-point circles of ABC, BCD, CDA and DAB concur at the anticenter of the cyclic quadrilateral.

If n is a positive integer, the integers between 1 and n − 1 which are coprime to n ( or equivalently, the congruence classes coprime to n ) form a group with multiplication modulo n as the operation ; it is denoted by Z < sub > n </ sub >< sup >×</ sup > and is called the group of units modulo n or the group of primitive classes modulo n. As explained in the article multiplicative group of integers modulo n, this group is cyclic if and only if n is equal to 2, 4, p < sup > k </ sup >, or 2 p < sup > k </ sup > where p < sup > k </ sup > is a power of an odd prime number.

If also, the cyclic quadrilateral becomes a triangle and the formula is reduced to Heron's formula.

* If a cyclic quadrilateral has side lengths that form an arithmetic progression the quadrilateral is also ex-bicentric.

* If the opposite sides of a cyclic quadrilateral are extended to meet at E and F, then the internal angle bisectors of the angles at E and F are perpendicular.

* If a cyclic quadrilateral is also orthodiagonal, the distance from the circumcenter to any side equals half the length of the opposite side.

If you examine the multiples of 1 / 7, you can see that each is a cyclic permutation of these six digits:

* If π < sub > 1 </ sub >( M ) is virtually cyclic but not finite then the geometric structure on M is S < sup > 2 </ sup >× R, and M is compact.

* If π < sub > 1 </ sub >( M ) is virtually abelian but not virtually cyclic then the geometric structure on M is Euclidean, and M is compact.

* If π < sub > 1 </ sub >( M ) has an infinite normal cyclic subgroup but is not virtually solvable then the geometric structure on M is either H < sup > 2 </ sup >× R or the universal cover of SL ( 2, R ).

If it is compact, then the 2 geometries can be distinguished by whether or not π < sub > 1 </ sub >( M ) has a finite index subgroup that splits as a semidirect product of the normal cyclic subgroup and something else.

* If π < sub > 1 </ sub >( M ) has no infinite normal cyclic subgroup and is not virtually solvable then the geometric structure on M is hyperbolic, and M may be either compact or non-compact.

If the sequence b < sub > n </ sub > were periodic in n with period N, then it would be a cyclic convolution of length N, and the zero-padding would be for computational convenience only.

If the random starting point is 3. 6, then the houses selected are 4, 19, 35, 51, 66, 82, 98, and 113, where there are 3 cyclic intervals of 15 and 5 intervals of 16.

If A has a multiplicative identity 1, then it is immediate that the equivalence class ξ in the GNS Hilbert space H containing 1 is a cyclic vector for the above representation.

If ord < sub > n </ sub > a is actually equal to φ ( n ) and therefore as large as possible, then a is called a primitive root modulo n. This means that the group U ( n ) is cyclic and the residue class of a generates it.

If X is the plane with the origin missing, and G is the infinite cyclic group generated by ( x, y )→( 2x, y / 2 ) then this action is wandering but not properly discontinuous, and the quotient space is non-Hausdorff.

If u is a non-zero element of U, u · R = U ( where u · R is the cyclic submodule of U generated by u ).

The converse also holds: If the sum of the distances from a point in the interior of a quadrilateral to the sides is independent of the location of the point, then the quadrilateral is a parallelogram.

If the quadrilateral has vertices u, v, w, x, then p

# If then the quadrilateral is a parallelogram.

# If and sgn r = sgn ( Im p ), then the quadrilateral is a trapezoid.

If a is the side of the large square and θ is the angle between two opposing sides in each quadrilateral, then the quotient between the two areas is given by sec < sup > 2 </ sup > θ − 1.

If a quadrilateral is known to be a trapezoid, it is not necessary to check that the legs have the same length in order to know that it is an isosceles trapezoid ; any of the following properties also distinguishes an isosceles trapezoid from other trapezoids:

If both lines are not diameters, the we may extend the tangents drawn from each pole to produce a quadrilateral with the unit circle inscribed within it.

The Grashof condition for a four-bar linkage states: If the sum of the shortest and longest link of a planar quadrilateral linkage is less than or equal to the sum of the remaining two links, then the shortest link can rotate fully with respect to a neighboring link.

If the quadrilateral is given with its four vertices A, B, C, and D in order, then the theorem states that:

If multiple symmetry axes are present, they will necessarily intersect at a single point, and this is usually taken to be the center.

The parallel postulate: If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough.

# If two straight lines in a plane are crossed by another straight line ( called the transversal ), and the interior angles between the two lines and the transversal lying on one side of the transversal add up to less than two right angles, then on that side of the transversal, the two lines extended will intersect ( also called the parallel postulate ).

If no minimum wage is in place, workers and employers will continue to adjust the quantity of labor supplied according to price until the quantity of labor demanded is equal to the quantity of labor supplied, reaching equilibrium price, where the supply and demand curves intersect.

If the navigator draws two lines of position, and they intersect he must be at that position.

In particular, if S and T ( subgroups now ) intersect only in the identity, then every element of ST has a unique expression as a product st with s in S and t in T. If S and T also permute, then ST is a group, and is called a Zappa-Szep product.

If the quantity under the square root ( the discriminant ) is negative, then the ray does not intersect the sphere.

If we now restrict our attention to that portion of the cos ( ωt ) coefficient which varies linearly with V, and then ask ourselves, at what input voltage level, V, will the coefficients of the first and third order terms have equal magnitudes ( i. e., where the magnitudes intersect ), we find that this happens when

If the geometric diagram does not intersect major physical points in the image, the result is what Skinner calls " unanchored geometry.

If one or more of the indices is zero, it means that the planes do not intersect that axis ( i. e., the intercept is " at infinity ").

If S is compact but not closed, then it has an accumulation point a not in S. Consider a collection consisting of an open neighborhood N ( x ) for each x ∈ S, chosen small enough to not intersect some neighborhood V < sub > x </ sub > of a.

* If any altitude, say AD, is extended to intersect the circumcircle at P, so that AP is a chord of the circumcircle, then the foot D bisects segment HP:

If the six connectors that join any pair of orthocentric points are extended to six lines that intersect each other, they generate seven intersection points.

If an object's bounding volume does not intersect a volume higher in the tree, it cannot intersect any object below that node ( so they are all rejected very quickly ).

If two chords of a circle AC and BD intersect at X, then the four points A, B, C, D are concyclic if and only if

If the extensions of opposite sides a and c intersect at an angle ', then

If the angle bisectors to angles A and B intersect at P, and the angle bisectors to angles C and D intersect at Q, then

If the line does not intersect the circle, there are no solutions.

:: Loopholes: If closed timelike curves exist, then timelike curves don't have to intersect the partial Cauchy surface.

If the Cauchy surface were compact, i. e. space is compact, the null geodesic generators of the boundary can intersect everywhere because they can intersect on the other side of space.