A tetrahedron ("four faces") is the most simple of the polyhedra: its four vertices define four triangular faces assembled with six edges.
The regular tetrahedron and the regular rightangled tetrahedron can easily be obtained from a cube.
a regular tetrahedron  a regular trirightangled tetrahedron  a diamond (compound of these two regular pyramids) 
Schläflis' birectangular tetrahedron (four rectangular faces) 
We can do so using a rectangular parallelepiped to obtain other types of tetrahedra. 

an equifacial tetrahedron (disphenoid) 
a trirightangled tetrahedron  an equifacial tetrahedron with isosceles faces (tetragonal disphenoid) 
an other tetrahedron with rightangled faces (non equifacial) 
Here are two other interesting tetrahedra defined in a cube (1/24 of the cube's volume):
• if we cut out the first (blue) along each of the twelve edges of a cube it remains a Kepler star (see puzzles), • 6 of the second (magenta) can build an oblate rhombohedron, 8 an octahedron (longest edge in common), 24 a cube (two along each edge), 6×8=48 a rhombic dodecahedron (six octahedra with a common vertex); twelve octahedra with a common vertex build a stellated rhombic dodecahedron. 
The four perpendiculars to the faces' planes and going through the centres of the circumcircles of the faces converge at the centre of the tetrahedron's circumsphere.
The center I of the sphere inscribed in a tetrahedron (ABCD) verifies I = (Sa/S)A+(Sb/S)B+(Sc/S)C+(Sd/S)D where Sa, Sb, Sc and Sd are the areas of the faces opposite to A, B, C and D respectively, and S=Sa+Sb+Sc+Sd ; its radius r verifies r=3V/S (by decomposing the tetrahedron into four tetrahedra with common vertex I we get V=Sr/3 ) and 1/r = 1/h_{1}+1/h_{2}+1/h_{3}+1/h_{4} where the h_{i} are the altitudes of the tetrahedron and V its volume. In a triangle ABC the incenter I verifies I = (a/p)A+(b/p)B+(c/p)C where a, b and c are the lengths of the sides opposite to A, B and C respectively, and p=a+b+c the perimeter of the triangle; its radius r verifies r=2S/p and 1/r = 1/h_{1}+1/h_{2}+1/h_{3} where the h_{i} are the altitudes of the triangle and S its area. The computation of the centers and the radii is not easy for a tetrahedron without special properties. Of course in the case of the regular tetrahedron things are simple (see exercise at the end of the page). Remarks: besides the insphere there are also four exspheres, like there are three excircles in a triangle. Only the Crelle's tetrahedra have a midsphere (tangent to the six edges). This figure may be dynamically modified by moving three of the tetrahedron's vertices (red points) with the mouse pointer. 
The centre of gravity (centroid or isobarycentre of the vertices) is the point of concurrency of the four segments joining each vertex to the centre of gravity of the opposite face; it lays on the quarter of each of these segments, starting from the face.
Generally a tetrahedron has no orthocentre (see orthocentric tetrahedra).
The volume of ABCD is the sixth of the volume of the parallelepiped constructed from the edges AB, AC and AD (half base area, same height). Using the coordinates of the vertices A(a_{x},a_{y},a_{z}), B(b_{x},b_{y},b_{z}), C(c_{x},c_{y},c_{z}) and D(d_{x},d_{y},d_{z}), it expresses itself with the help of a determinant:

triangle's area by Heron: sides a, b and c, A²=s(sa)(sb)(sc) where s=(a+b+c)/2 is the halfperimeter.
area of a convex quadrilateral: sides a, b, c and d, A²=(sa)(sb)(sc)(sd)abcd×cos(S)²
where s=(a+b+c+d)/2 is the halfperimeter and S half the sum of two opposite angles; for cyclic quadrilaterals cos(S)=0
proof: the computation is easy in the case of a regular tetrahedron inscribed in a cube with edge 1 (each of the four trirightangled tetrahedra has a volume of 1/6; thus it remains 1/3 for the regular tetrahedron). Generalizing is easy: the four pyramids have the same volume (same height, that of the parallelepiped, and bases with same area, half of that of the parallelepiped's base). The edges of the tetrahedron are six of the twelve diagonals of the parallelepiped's faces.
The "f" key is a switch to display the faces or only the edges. 
In every quadrilateral (even non convex or crossed) the midpoints of the sides are the vertices of a parallelogram.
If the diagonals of the quadrilateral have same length (resp. are perpendicular) then we get a rhombus (resp. a rectangle). Four points define three quadrilaterals (on the left diagram one is convex and the two others are crossed) whose three Varignon's parallelograms have same centre (the common midpoint of their diagonals). In the second configuration the three quadrilaterals are non convex and non crossed. In every quadrilateral the three segments joining the midpoints of opposite sides and the midpoints of the diagonals have same midpoint (Varignon's point). The theorem remains true in space (right diagram); the four non coplanar points are then the vertices of a tetrahedron; we can see on the applet above that the common midpoint is the centre of the parallelepiped.

Here we really change dimension (move from the plane to the space):
• quadrilateral → octahedron • midpoint of a side → centroid of a face • parallelogram → parallelepiped This figure may be dynamically modified by moving the vertices of the tetrahedron (yellow points) with the mouse pointer. The "f" key is a switch to display the faces of the octahedron. Remark: The octahedron doesn't need to be convex: we may consider a crossed quadrilateral as two triangles which intersect on the same side of their common base; likewise a "crossed" octahedron may be seen as two "pyramids" which intersect on the same side of their common "base" (in fact it's a set of four non necessary coplanar points). By mowing the vertices of the octahedron we can test those configurations. 
proof 1 (Vladimir Dubrovsky):
Let the midpoints M and N be taken on edges AB and CD, and let the other two vertices of the section (in light blue) be K (on AC) and L (on BD). A parallel projection of the whole construction along MN gives a parallelogram ACBD with center M=N in which AK/AC = BL/BD = k. One "half" of (ABCD) falls into the 4sided pyramid (CKMLN) and the tetrahedron (LBCM), the other one into (DKMLN) and (KADM), and vol(CKMLN)=vol(DKMLN) because they have same base and equal heights vol(BCML) = vol (ADMK) = (k/2)×vol(ABCD) because bases(ABCD)/2 and heights(ABCD)×k If K=A and L=B (or K=C and L=D) the two "halves" are tetrahedra with same volume: height(ABCD) and base(ABCD)/2 
projection  
proof 2 (Vladimir Dubrovsky) :
Consider the sections of (ABCD) parallel to AB and CD. They are parallelograms and the section cuts each of them into two equal symmetric parts (see the projection above). Then the equality of the volumes of the two pieces follows by "Cavalieri's Principle." proof 3 (Georges Lion) :
generalization: If a plane divides two opposite edges of a tetrahedron in a given ratio, then it divides the volume of the tetrahedron in the same ratio. (AltshillerCourt 1979, page 89) 
In fact we have a sufficient condition much more weak (Georges Lion).
If one of the altitude is secant with two others, not necessarily a priori at the same point, then the tetrahedron is orthocentric. proof: For line AB to be orthogonal to line CD it is necessary and sufficient that h(A), altitude going through A, is secant with h(B). Let us notice that generally h(A) and h(B) are orthogonal to CD without being parallels. We have then: h(A) and h(B) are secant <=> h(A) and h(B) are coplanar <=> there is a plane going through A and B and perpendicular to CD. If this condition is carried out the edges AB and CD are orthogonal and so are the edges AC and BD, thus the tetrahedron is orthocentric. Remark: if h(A) and h(B) are secant then so are h(C) and h(D) but the tetrahedron ABCD is not necessarily orthocentric. 
For any tetrahedron ABCD V=(1/6)×AB×CD×h×sin(x) where x is then angle (AB,CD)
The angle of two non coplanar lines is the one of their parallels going through some point (on the construction H is such a point). reminder: construction of the common perpendicular Δ to two non coplanar lines D and D' Let D" be the orthogonal projection of D' on the plane P going through D and parallel to D', and H its intersection point with D; Δ is the line perpendicular to P going through H. This drawing is interactive: you may modify the tetrahedron by moving the red points B, C and D, and thus you can verify that the common perpendicular can be outside of the tetrahedron. 
a+a'=b+b'=c+c' where l and l' are the length of two opposite edges
example: the "4ball tetrahedra"
See also Soddy's "kissing spheres". 
The net of the regular tetrahedron is well known (an equilateral triangle with its midpoint triangle); the net of the regular trirectangle tetrahedron can be easily deduced.  
Here is the second net of the regular tetrahedron:  
A triangle folded along the sides of its midpoints triangle is the net of an equifacial tetrahedron (the triangle must have three acute angles).
Three folds a square into the net of a trirightangled tetrahedron. A square may also be the net of an equifacial tetrahedron.  
If we pull one lateral face of a pyramid down on the plane of its base, we notice that the altitude of this face swings around its foot. We deduce the construction of a net of a tetrahedron:
We choose the base triangle and a point S (projection of the summit) which may be chosen on a side or outside of the base triangle. Each lateral triangle has an altitude coming through S, on which lays its third vertex. We choose one of them (S' for example, with HS'>HS) and deduce the two others by transferring the lengths of the sides. We may choose the height SS'=h of the tetrahedron by constructing the rightangled triangle HSS'. 
If we cut a tetrahedron SABC using a plane parallel to the face ABC, for example at midhigh, we get a pentahedron (truncated pyramid with three trapezoid lateral faces), whose volume is 7/8 of the tetrahedron's volume.
The altitudes of the three trapeziums are obviously not equal, but there is a direction in the space according to which they seem to be. We may search it experimentally by moving the solid. 

>Let's switch from the space to the plane by projecting the summit S to S' on the plane (ABC). The section at midhigh is then a triangle A'B'C', and the midpoint theorem in a triangle proves that A'B'C' is a reduction of ABC (for a reduction ratio different from 1/2 we use Thales theorem, or an homothecy with centre S'). In order to have three trapeziums with same altitude, it is necessary and sufficient for each vertex of A'B'C' to be equidistant from two sides of ABC, thus laying on the bisector of the sector. Finally, S' has to be the incenter I of ABC. The searched direction is (SI). 
references:  Géométrie de l'espace et du plan (in French) by Yvonne and René Sortais (published by Hermann  1988) pages 305336
Building space fillers with tetrahedra ("mites")&tninsp; by Eduard Bobik 
See also the pages devoted to the decompositions of the parallelepiped, to the tetrahedral symmetries and to the kaleidocycles.
April 1999 updated 03012014 