
Contents of this page
 Parallel lines and congruent angles  elementary geometrical facts
 Squares, Rectangles  area and perimeter facts
 Quadrilateral and triangle area theorems  Bretschneider, Brahmagupta, Heron, Pick's,
 Cyclic quadrilateral theorems  circumradius, Ptolemy's, Isosceles trapezoid, puzzle
 Concurrence theorems  Carnot, Ceva, Brianchon, Pascal, Pappus
 Triangle theorems  4 triangle centers, right triangle median/bisector/altitude, crossed ladder, isogonal conjugates, Urquhart, puzzles
 Similar figures aligned with edges of a polygon  . . . . . .
 Circles, ellipses  Inscribed angle, Intersecting chords, Descartes, Ellipse reflection
 Inversion circle  Inversion circle, Pappus chain of circles in an arbelos
 Miscellaneous  puzzles and other factoids that don't fit anywhere else
 Projective geometry  point/line duality, point plane duality
Parallel lines and Congruent Angles
Corresponding Angles Postulate: If two coplanar lines are cut by a transversal so that two corresponding angles have the same measure, then those lines are parallel. (converse of Parallel Lines Postulate).
Parallel Lines Postulate: If two lines are parallel and are cut by a transversal, corresponding angles have the same measure. (converse of Corresponding Angles Postulate).
Linear Pair Theorem: If two angles form a linear pair, then they are supplementary. (Proof: given linear pair <ABD and <DBC, BA and BC are opposite rays, so m<ABC=180 degrees. m<ABD+m<DBC=m<ABC (angle addition assumption), m<ABD+m<DBC=180 degrees (substitution), so <ABD and <DBC are supplementary by definition of supplementary.)
Vertical Angle Theorem: If two angles are vertical angles, then they have equal measures. (Proof: given vertical angles <BEA and <CED, <1 and <3 are a linear pair and thus supplementary (def of linear pair, linear pair theorem), <2 and <3 are a linear pair, and thus supplementary. m<1+m<3=180, m<2+m<3=180 (def of supplementary angles). m<1=m<2 (substitution property, subtraction property).)
Two angles supplementary to the same angle are congruent to each other: Corollary to the Vertical Angle Theorem; the proof is identical to that given for the Vertical Angle Theorem.
Two Pependiculars Theorem: If two coplanar lines a and b are each perpendicular to the same line, then they are parallel to each other. (Proof: special case of the Corresponding Angles Postulate)
Perpendicular to Parallels Theorem: In a plane, if a line is perpendicular to one of two parallel lines, it's perpendicular to the other. (Proof: special case of the Parallel Lines Postulate)
Alternate Exterior Angles Theorem (and its converse): Given two lines cut by a transversal, the lines are parallel iff the alternate exterior angles are congruent. (Proof of onlyif direction: <1=<5 by Parallel Lines Postulate; <5=<7 by Vertical Angle Theorem; <1=<7 by Transitive Property of Congruence of Angles)
Alternate Interior Angles Theorem (and its converse): Given two lines cut by a transversal, the lines are parallel iff the alternate interior angles are congruent. (Proof of onlyif direction: <2=<6 by Parallel Lines Postulate, <4=<2 by Vertical Angle Theorem, <4=<6 by Transitive Property of Congruence of Angles)
Interior Angles Theorem (and its converse): Given two lines cut by a transversal, the lines are parallel iff the interior angles on the same side of the transversal are supplementary. (Proof of onlyif direction: <1=<5 by Parallel Lines Postulate; <1 and <4 are supplementary by Linear Pair Theorem; m<1+m<4=180 by def of supplementary, m<5+m<4=180 by substitution, and m<5 and m<4 are supplementary by def of supplementary)
Right Angle Congruence Theorem Proof: All right angles are congruent. (Proof: given arbitrary right angles <1 and <2, m<1=90 degrees by def of right angle, m<2=90 degrees, m<1=m<2 by transitive property of equality, <1=<2 by def of angle congruence.
Puzzle: Area of the smaller square  using the Pythagorean Theorem to calculate the area of a figure
Rectangle Perimeter  Proof that a square has the smallest perimeter of any rectangle with a given area
Quadrilateral and triangle area theorems
Bretschneider's Formula for the area of a quadrilateral with sides of length a, b, c, d and opposite interior angles A and C (it turns out not to matter which two opposite corners are selected). Then its area, K = sqrt[(sa)(sb)(sc)(sd)  abcd cos^{2}([A+C]/2)]. If ABCD is cyclic, [A+C]/2 = 90� and this reduces to the Brahmagupta formula for the area of a cyclic quadrilateral (of which Heron's formula for a triangle is a yetmorespecial case).
Heron's Formula for the area of a triangle, K = sqrt(s(sc)(sb)(sa)), where s=(a+b+c)/2, the semiperimeter. It is proved using the law of cosines.
Pick's Theorem for the area of a lattice polygon, Area(P) = i + (b/2)  1, where i is the number of interior lattice points in P and b is the number of boundary lattice points on P.
Fundamental Theorem of Directly Similar Figures (FTDSF)  if the lines connecting the corresponding vertices of two directly similar polygons are divided in equal ratios, then the resulting polygon is directly similar to the given two polygons.
The FinslerHadwiger Theorem  Squares ABCD and PQRS joined at a corner D=P; the figure formed by joining the midpoints of AC, CQ, QS, and SA is a square. Special case of FTDSF. This factoid serves as the springboard for a proof of this factoid:
The figure formed by joining the side midpoints of a quadrilateral is a parallelogram is proved using vectors, along with the fact that the area of a quadrilateral is half the vector cross product of its diagonals. Moreover the area of the parallelogram is half that of the quadrilateral.
Cyclic quadrilateral with consecutive sides a,b,c,d, diagonals p,q, and semiperimeter s:
circumradius R = (1/4)sqrt((ac+bd)(ad+bc)(ab+cd)/((sa)(sb)(sc)(sd))).
Ptolemy's Theorem for cyclic quadrilateral: ac+bd = pq. Ptolemy's inequality for a convex (not necessarily cyclic) quadrilateral: ac+bd ≥ pq with equality iff cyclic.
Butterfly Theorem: given a chord PQ with midpoint M, and two other chords AB and CD that intersect at M, then M is the midpoint of the intersections with PQ by AD and CB. Additionally (or alternatively), if AB and CD are considered the diagonals of cyclic quadrilateral CADB, then opposite sides of CADB intersect PQ at points equidistant from M.
Isosceles Trapezoid Proofs: proofs of various factoids about isosceles trapezoids.
Puzzle: Rectangle and Cyclic Quadrilateral, and its solution: interesting properties of a particular class of cyclic quadrilateral, given by the coordinates of its vertices.
(Note: 3 points are said to be "concurrent" if they are collinear; 3 lines concur if they meet in a point)
Carnot's Theorem  Let points A', B', and C' be located on the sides BC, AC, and respectively AB of ABC. The perpendiculars to the sides of the triangle at points A', B', and C' are concurrent iff AC'^{2}  BC'^{2} + BA'^{2}  CA'^{2} + CB'^{2}  AB'^{2} = 0.
Ceva's Theorem  three arbitrary cevians AA', BB' and CC' of triangle ABC concur iff AC'/C'B � BA'/A'C � CB'/B'A = 1
van Aubel's second theorem: if triangle A'B'C' is the Cevian triangle of a point P, then AP/PA' = AB'/B'C + AC'/C'B generalizes the fact that the centroid of a triangle cuts each Cevian segment through it in the ratio 2:1. (See also van Aubel's theorem)
Brianchon's Hexagon Theorem: in a hexagon circumscribed about a conic, the major diagonals, i.e. the diagonals joining vertices with the opposite ones, are concurrent. (Its dual is Pascal's Theorem)
Pascal's Hexagon Theorem: if a hexagon is inscribed in a conic, then the three points at which the pairs of opposite sides meet lie on a straight line. (Its dual is Brianchon's Theorem)
Pappus' Theorem: Let three points A, B, C be incident to* a single straight line and another three points a,b,c incident to another straight line. Then three pairwise intersections 1 = BcbC, 2 = AcaC, and 3 = AbaB are incident to a (third) straight line. (*See the note under "projective geometry", below, regarding the use of "incident to".)
Cross Ratio Theorem: Given four collinear points A,B,C,D and four coplanar and concurrent lines a,b,c,d, the point cross ratio (ABCD) = CA/CB : DA/DB equals the line crossratio (abcd) = sin(cMa)/sin(cMb) : sin(dMa)/sin(dMb), where line segments and angles are considered "signed".
Four key Triangle Centers  Centroid, Circumcenter, Incenter (with the Angle Bisector Theorem for good measure), and Orthocenter.
Apollonius' theorem  in triangle ABC, if point D on BC divides BC in the ratio n:m then mAB^{2} + nAC^{2} = mBD^{2} + nDC^{2} + (m + n)AD^{2}.
Right triangle median, altitude, bisector: The median and altitude of triangle ABC are reflections about the bisector of angle C iff ABC is a right triangle.
Extended Crossed Ladder Theorem  When a pair of ladders are crossed inside a triangle of height c, with the intersection of ladders occurring at height f, and the tops of the ladders against the sides of the triangle at heights d and e, then 1/c + 1/f = 1/d + 1/e. If the walls the ladders are leaning on are parallel, then 1/f is taken as zero, for the "traditional" crossed ladder theorem.
SteinerLehmus theorem  Any triangle with two angle bisectors of equal length is isosceles.
The Isogonal Conjugates of a point, P, is a point P' at the intersection of the cevians you get by reflecting each cevian about the bisector of the angle of the original triangle. Ceva's Theorem, mentioned on this page under "concurrence" proves these reflected cevians concur.
Trisected Triangle Puzzle: Proof of a puzzle question in which one angle of a triangle is trisected, another is bisected; a length is given, another is sought. Proof uses several of these triangle theorems.
Urquhart's Theorem: If lines OA and OB intersect at O, A' is on OA, B' is on OB, AB' and A'B intersect at O'; Then OA + AO' = OB + BO' iff OA' + A'O' = OB' + B'O'.
Equilateral Puzzle: the distinct complex numbers a, b and c represent the vertices of an equilateral triangle if and only if a�+b�+c�abbcca=0
See Napolean's Theorem, below.
Similar figures aligned with edges of a polygon
Van Aubel's Theorem: Line segments connecting centers of squares on opposite sides of a quadrilateral are perpendicular and equal in length. (See also van Aubel's second theorem.)
Thebault's Problem 1: Line segments connecting centers of squares on sequential sides of a parallelogram form a square. This is a special case of van Aubel's Theorem. Problem 2: The vertices of equilateral triangles constructed on the outside (or inside) of two consecutive sides of a square, together with the "untouched" vertex of the square form an equilateral triangle.
Napoleon's Theorem: if equilateral triangles are constructed on the sides of any triangle (all outward or all inward), the centers of those equilateral triangles themselves form an equilateral triangle.
The Inscribed Angle Property: (also known as the Central Angle Theorem): The measure of an angle inscribed in a circle is half the measure of the arc it intercepts.
An extension to this property is that Intersecting Chords form an angle equal to the average of the arcs they intercept.
The Intersecting Chords Theorem says the products of the two segments of chords cut by their point of intersection are equal.
Descartes' Circle Theorem: Given four mutually tangent circles whose curvatures (curvature is the reciprocal of radius) are k_{1}, k_{2}, k_{3}, and k_{4}, (k_{1}+k_{2}+k_{3}+k_{4})^{2} = 2(k_{1}+k_{2}+k_{3}+k_{4}).
Mutual Radial Tangents Puzzle: Given two circles, each external to the other, draw two lines (extended radii) from the center of each circle that are tangent to the other circle, then prove the chords intercepted by the extended radii in each circle are equal in length.
The Ellipse Reflection Property  If the interior of ellipse is silvered to produce a mirror, rays originated at ellipse's focus are reflected to the other focus of the ellipse.
Pappus Chain of circles in an arbelos  Construction of Pappus Chain using Inversion Circle
Puzzle: Three Circles and Triangle, and its solution: an ancient Sangaku problem that seems to defy simple geometric solution.
Projective Geometry: A geometry which begins with the ordinary points, lines, and planes of Euclidean plane geometry, and adds an ideal plane, consisting of ideal lines, which, in turn contain ideal points, which are the intersections of parallel lines and planes.
In projective plane geometry, points and lines are considered "duals" of one another. Two lines always determine (contain) exactly one point; two points always determine exactly one line.
In projective space geometry, points and planes are considered duals of one another. Three noncollinear points determine a plane; three noncollinear planes determine a point.
The axioms of projective geometry are duals of one another as well, which means the words "point" and "line" can be interchanged in any axiom to get another axiom. Even more startling is that any proof using these axioms, or derived from other proofs using the axioms can also be changed in the same way to prove its dual.
* To make the terminology easier to convert to its dual terminology, the word "incident to" is used both of a point lying on a line, and of a line passing through a point. For example, three collinear points are said to be incident to a particular line, and three concurrent lines are said to be incident to a particular point.
Cuttheknot: Projective Geometry
Elementary theorems of geometry: regentsprep.org/regents/mathb/1b/theorems.htm
IMO Interactive Math Online: The Basic Postulates & Theorems of Geometry
The webmaster and author of this Math Help site is Graeme McRae.