Target Audience: Anyone who has met triangles and circles at school and wants to see the subject built rather than asserted.
Prerequisites: Tier 0 — in particular Proof Techniques. Geometry is where proof was invented, and almost every result below is proved by direct argument or by contradiction.
This page has one destination. It starts with undefined terms and five postulates, and ends at the AA Similarity Theorem — the result that lets you conclude two triangles have proportional sides from nothing but their angles. Everything between is the chain that earns it.
The pivot is Section 4. Congruence is what lets one figure be transferred onto another, and every later theorem leans on it. The triangle explorer there solves a triangle from the data you give it and tells you whether that data pins the triangle down — including the case that does not, which is exactly why the list of congruence criteria is SAS, ASA and SSS and not four items long.
Second unit of Tier 1. Each section names the proof page that carries the full argument; this page is the ladder between them, not a replacement for them.
Euclidean geometry is the oldest worked example of the method described in Proof Techniques: fix a small number of assumptions, then derive everything else. Around 300 BCE Euclid set out the assumptions explicitly and proved some 465 propositions from them. That structure — not the particular facts about triangles — is what made the Elements the model for mathematics for two thousand years.
Not everything can be defined. Any definition uses earlier words, so the chain has to start somewhere. Euclidean geometry starts with point, line and plane left undefined; they are constrained only by the axioms that mention them, never described.
This is the same move the Sets, Relations and Functions page makes with "set". A term that everything else is built from cannot itself be built.
Euclid's five postulates, in modern phrasing:
The first four are short and self-evident; the fifth is neither. For centuries mathematicians tried to prove it from the other four, and every attempt failed. In the 1820s Bolyai and Lobachevsky established why: it is genuinely independent. Deny it and you get a different, perfectly consistent geometry.
So the fifth postulate is a choice, and "Euclidean" names the choice. Everything on this page that depends on it — the angle sum being exactly \(180^\circ\), similarity behaving as it does — is false in hyperbolic geometry. That is worth knowing before you meet the claim that these facts are simply true of space.
An angle is the figure formed by two rays sharing an endpoint, called the vertex. Its measure is how far one ray must turn to reach the other, in degrees, with a straight angle measuring \(180^\circ\) and a right angle \(90^\circ\).
The first genuine theorem is about that last pair, and it is short enough to give in full.
Two lines cross, making four angles. Call two adjacent ones \(\alpha\) and \(\beta\), and let \(\gamma\) be opposite \(\alpha\).
\(\alpha\) and \(\beta\) together form a straight angle, so \(\alpha + \beta = 180^\circ\). So do \(\beta\) and \(\gamma\), so \(\beta + \gamma = 180^\circ\). Subtracting, \(\alpha = \gamma\). \(\blacksquare\)
Two lines, one subtraction. This is the flavour of the whole subject: the content is in choosing which two facts to set side by side.
Two lines in a plane are parallel if they never meet. A third line crossing both is a transversal, and it creates eight angles whose relationships carry most of the work in elementary geometry.
When the two lines are parallel, alternate interior angles are equal, corresponding angles are equal, and co-interior angles are supplementary. The first of these is the one everything else is derived from, and it is where the Parallel Postulate enters.
The Alternate Interior Angles Theorem proves both directions: parallel lines give equal alternate interior angles, and equal alternate interior angles force the lines to be parallel. That converse is the tool for proving lines parallel, and it is used in the next two sections.
Two triangles are congruent when one can be placed exactly on the other: all three sides equal and all three angles equal, in matching order. Checking all six parts would be tedious, and the congruence criteria say you never have to — three well-chosen parts are enough.
The word between is doing real work in the first two. Drop it from SAS — take two sides and an angle that is not between them, the pattern called SSA — and the criterion fails. The data no longer determines the triangle.
Choose a criterion. The explorer solves the triangle from that data by the laws of sines and cosines, reports every side and angle, and says how many triangles fit. Watch what happens on the three SSA rows.
Covered in Section 7 — come back once you have read it. Scale a 6-7-8 triangle and watch which quantities move.
Every value and every vertex is computed live from the data shown — nothing is tabulated in advance.
With \(a = 7\), \(b = 10\) and \(A = 30^\circ\), the explorer finds two triangles: one with \(B \approx 45.58^\circ\) and one with \(B \approx 134.42^\circ\). Both genuinely have those two sides and that angle. Since \(\sin B = \sin(180^\circ - B)\), the law of sines cannot distinguish them.
Shorten \(a\) to \(3\) and no triangle closes at all; lengthen it to \(10\) and the second solution leaves no room for a third angle, so exactly one survives. Zero, one or two — a criterion has to give one, every time, which is why SSA is not on the list.
A triangle is isosceles when two of its sides are equal, and equilateral when all three are. The first theorem proved with the congruence criteria is about the isosceles case, and it is old enough to have its own name: pons asinorum, the bridge of asses.
If two sides of a triangle are equal, the angles opposite them are equal.
The slick proof compares the triangle with itself, taken in the other order: in triangle \(ABC\) with \(AB = AC\), the correspondence sending \(A \to A\), \(B \to C\) and \(C \to B\) matches side to equal side and the angle at \(A\) to itself. By SAS the triangle is congruent to its own mirror image, so the angle at \(B\) equals the angle at \(C\). \(\blacksquare\)
The Isosceles Triangle Theorem gives the argument in full, along with its converse — equal angles force equal sides — which is what makes "isosceles" and "two equal angles" interchangeable in later proofs.
The angles of any triangle total \(180^\circ\). The proof is one construction plus Section 3: through one vertex draw the line parallel to the opposite side, and the three angles at that vertex — two of them alternate interior angles with the triangle's own — make a straight angle.
That construction needs a parallel through a point, and needs it to be the only one. Without the fifth postulate the theorem is false: in hyperbolic geometry every triangle has angle sum strictly less than \(180^\circ\), and the shortfall grows with area. The full argument is on The Triangle Angle Sum Theorem.
Extend one side past a vertex and the angle outside is the exterior angle there. It equals the sum of the two remote interior angles — immediate from the angle sum, since both quantities are what remains after subtracting the adjacent interior angle from \(180^\circ\).
The Exterior Angle Theorem. Its weaker form — an exterior angle exceeds either remote interior angle — is proved without the Parallel Postulate, which is why it survives in geometries where the angle sum does not.
A triangle has at most one angle that is \(90^\circ\) or more: two would already reach \(180^\circ\), leaving nothing for the third. So "the largest angle" is always well defined, and the side opposite it is always the longest — the fact behind The Triangle Inequality, which the explorer above uses to reject impossible SSS data.
Two triangles are similar when they have the same shape but not necessarily the same size: corresponding angles equal, corresponding sides in a constant ratio. Congruence is the special case where that ratio is \(1\).
If two angles of one triangle equal two angles of another, the triangles are similar.
Two angles suffice because the third comes free from the angle sum. So the whole shape of a triangle is pinned down by two numbers, and the only remaining freedom is scale.
This is the destination of the page. Everything before it was needed: congruence to transfer figures, the parallel-line angles to produce equal angles from a construction, and the angle sum to get the third angle for free. The proof runs by cutting the larger triangle down to a copy of the smaller one and showing the cut is parallel to the third side, which makes the sides proportional.
The AA Similarity Theorem carries the full proof, and its step-by-step walkthrough takes the construction one stage at a time.
Use the slider in the explorer to see what similarity fixes and what it does not: angles and side ratios hold at every scale, lengths multiply by \(k\), and area by \(k^2\).
Doubling every side multiplies the area by \(4\), not \(2\). The explorer's area row shows this directly. It is the single most common slip in similarity problems, and the reason a map at twice the scale needs four times the paper.
In a right triangle with legs \(a\) and \(b\) and hypotenuse \(c\),
The proof that fits this page drops the altitude from the right angle to the hypotenuse. That splits the triangle into two smaller ones, each similar to the original by AA — every one of the three shares an angle with the original and has a right angle. Writing down the resulting proportions and adding them gives the theorem. Similarity, from the previous section, is the whole engine.
The Pythagorean Theorem, and its converse — if \(a^2 + b^2 = c^2\) then the triangle is right-angled — which is what licenses using the equation as a test rather than only as a formula. The converse is proved with SSS, tying it back to Section 4.
Solving the SSS data \(3, 4, 5\) returns angles \(36.87^\circ\), \(53.13^\circ\) and exactly \(90^\circ\) — the converse in action, since nothing in the input said the triangle was right-angled.
A circle is the set of points at a fixed distance — the radius — from a fixed centre. A chord joins two points on it; a chord through the centre is a diameter. An angle with its vertex at the centre is a central angle; one with its vertex on the circle is an inscribed angle.
An inscribed angle is half the central angle standing on the same arc.
The proof is a clean use of Section 5: join the vertex to the centre, and the radii make isosceles triangles whose base angles are equal. Pons asinorum, applied twice, does all the work.
Two consequences follow at once. All inscribed angles on the same arc are equal, since each is half the same central angle. And when the arc is a semicircle the central angle is \(180^\circ\), so the inscribed angle is \(90^\circ\) — Thales' theorem: any angle inscribed in a semicircle is a right angle.
The Inscribed Angle Theorem and Thales proves the general statement by cases according to where the centre falls relative to the angle, then derives Thales as the semicircle case.
A quadrilateral is cyclic when all four vertices lie on one circle. For such a quadrilateral \(ABCD\), Ptolemy's theorem states that the product of the diagonals equals the sum of the products of the two pairs of opposite sides:
This is the natural end point of the page, because its proof uses nearly everything above: a construction, the inscribed angle theorem to produce equal angles, AA similarity to turn those into proportions, and algebra to add the proportions up.
Applied to a rectangle it collapses to the Pythagorean theorem. Applied to points on a unit circle it yields the sine and cosine addition formulas — the identities derived independently in the trigonometric identities tutorial. That is a genuine bridge between this unit and Tier 1's trigonometry.
Ptolemy's Theorem, including the inequality form for non-cyclic quadrilaterals.
The chain, in the order it was built:
Thirteen proof pages in the Geometry category now have a path leading to them. The two that were hardest to reach cold — AA Similarity and Ptolemy — sit at the end of that path rather than at the start.
Next comes trigonometry, which is similarity with the ratios given names. The fact that the ratio of two sides depends only on the angles — not on the size of the triangle — is Section 7, and it is the entire reason the sine of an angle is well defined.
| Fact | Statement | Proof page |
|---|---|---|
| Vertical angles | Opposite angles at a crossing are equal | Section 2 above |
| Alternate interior angles | Equal if and only if the lines are parallel | Alternate Interior Angles |
| SAS / ASA / SSS | Three parts determine a triangle up to congruence | SAS, ASA, SSS |
| SSA | Not a criterion — admits 0, 1 or 2 triangles | Explorer, Section 4 |
| Pons asinorum | Equal sides give equal opposite angles, and conversely | Isosceles Triangle |
| Angle sum | \(A + B + C = 180^\circ\) | Triangle Angle Sum |
| Exterior angle | Equals the sum of the two remote interior angles | Exterior Angle |
| Triangle inequality | Each side is shorter than the sum of the other two | Triangle Inequality |
| AA similarity | Two equal angles give the same shape; sides in constant ratio | AA Similarity |
| Scaling | Lengths by \(k\), areas by \(k^2\), angles unchanged | Explorer, Section 4 |
| Pythagoras | \(a^2 + b^2 = c^2\), and the converse | Theorem, Converse |
| Inscribed angle | Half the central angle on the same arc; Thales is the semicircle case | Inscribed Angle and Thales |
| Ptolemy | \(AC \cdot BD = AB \cdot CD + BC \cdot AD\) for a cyclic quadrilateral | Ptolemy |
| Law of cosines | \(a^2 = b^2 + c^2 - 2bc\cos A\) — Pythagoras with a correction term | Used by the explorer |
| Law of sines | \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\) | Used by the explorer |