The Timpson Lyceum · Geometry (2024)

Euclid's Elements & Constructions – Lecture 9 (Geometry)

Overview

This lecture develops the ordering theorems that flow from the triangle congruence criteria (SAS, ASA, SSS), then steps back to place all of this work in its historical home: Euclid's Elements and straightedge-and-compass (ruler-and-compass) construction. We first prove the Exterior Angle Theorem, the relationship between a triangle's side lengths and its opposite angles, and the Triangle Inequality (with its algebraic counterpart). We then reconstruct several of Euclid's earliest propositions to see why drawing a figure with these two tools is a proof: each tool action is logically equivalent to invoking one of Euclid's five postulates. The lecture closes with the three classical construction problems the Greeks could never solve — trisecting an angle, duplicating the cube, and squaring the circle — all provably impossible with compass and straightedge alone.

Key Concepts

Exterior angle and opposite interior angles. An angle that is adjacent (next to) and supplementary (together forming a straight angle) to an interior angle of a triangle is an exterior angle of that triangle. The two angles of the triangle not adjacent to it are its opposite interior angles. A triangle has many exterior angles.

Exterior Angle Theorem. An exterior angle of a triangle is strictly greater than either of its opposite interior angles. The proof is by contradiction: it cannot be congruent to an opposite interior angle (that would force three points to be collinear, so two distinct lines would meet twice), and it cannot be smaller (that case reduces back to the congruent case inside a smaller triangle).

Sides and opposite angles. In any triangle the larger side lies opposite the larger angle, and vice versa. Stated for $\triangle ABC$ as an "if and only if":

If $AB > AC$ then $\angle C > \angle B$ (the angle opposite the longer side is larger).
If $\angle C > \angle B$ then $AB > AC$ .

The forward direction uses an isosceles cut and the Exterior Angle Theorem; the converse is proved by contradiction (the side cannot be equal — isosceles base angles would be equal — and cannot be smaller — that would reverse the inequality).

Triangle Inequality. In any triangle, the sum of two side lengths exceeds the third: $AB + BC > AC.$ Intuitively, the shortest path between two points is the straight segment joining them. For any three points (allowing the collinear, degenerate case) the relation relaxes to $AB + BC \ge AC$ , with equality exactly when $B$ lies on segment $AC$ .

Algebraic Triangle Inequality. With the absolute value defined piecewise, $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$ , two lemmas hold: $x \le |x|$ and $|x| = |{-x}|$ . From these, $|a + b| \le |a| + |b|.$ Equality holds when $a$ and $b$ share the same sign.

Polygonal path and polygon. Given ordered points $P_1, P_2, \dots, P_n$ , the polygonal path is the chain of segments $P_1P_2,\, P_2P_3,\, \dots,\, P_{n-1}P_n$ . If the last point coincides with the first, the figure closes into a polygon; the points are its vertices and the segments its sides. The length of a polygonal path is the sum of its segment lengths, and it is always at least the straight distance $P_1P_n$ (proved by repeatedly applying the Triangle Inequality to shortcut one vertex at a time).

Circle vocabulary. A circle with center $O$ and positive radius $r$ is the set of points $A$ with $OA = r$ . Points with $OA > r$ are exterior, those with $OA < r$ are interior. A radius is a segment from $O$ to a point on the circle; a diameter is a chord through $O$ , of length $2r$ . A chord joins two points on the circle, and the line through a chord is a secant (a tangent touches at only one point). A central angle has its vertex at $O$ ; an inscribed angle has its vertex on the circle and both sides meeting the circle at a second point (so its sides contain chords).

Euclid's foundations. Euclid's Elements builds all of geometry from just five postulates and five common notions:

Postulate 1: any two points can be joined by a segment (the straightedge).
Postulate 2: a segment can be extended indefinitely (the straightedge again).
Postulate 3: a center point and a radius determine a circle (the compass).
Postulate 4: all right angles are congruent.
Postulate 5 (parallel postulate): through a point not on a given line there is exactly one line that never meets it.

The common notions are more foundational than geometry — they are logic common to all fields: equality is transitive; equals added to equals are equal; equals subtracted from equals are equal; things that coincide are equal; and the whole is greater than any of its parts. Crucially, every compass-and-straightedge action in a drawing is logically equivalent to invoking one of these postulates, which is why constructing the figure is the same act as writing the proof.

The three classical impossible constructions. Using only compass and straightedge:

Trisecting an arbitrary angle is impossible (bisecting, by contrast, is easy and can be iterated to split into $4, 8, 16, \dots$ parts).
Duplicating the cube is impossible: doubling a unit cube's volume requires a side of length $\sqrt[3]{2}$ , which cannot be constructed (though $\sqrt{2}$ , the diagonal of a unit square, is trivial).
Squaring the circle — building a square with the same area as a given circle — is impossible.

These failures are not gaps in Euclid's reasoning but limits of his postulate set; a more rigorous modern axiom system can describe such objects even where compass-and-straightedge cannot build them.

Worked Example

Euclid's Proposition 1 — constructing an equilateral triangle on a given segment $AB$ . The construction draws two circles and joins an intersection point; each step is justified by a postulate, common notion, or the definition of a circle.

Draw circle centered at $A$ with radius $AB$ . (Postulate 3)
Draw circle centered at $B$ with radius $AB$ . (Postulate 3)
Let $C$ be a point where the two circles intersect. Join $AC$ and $CB$ . (Postulate 1)
Conclude all three sides are equal.

\begin{aligned} AC &\cong AB \quad \text{(def. of circle centered at } A) \\ BC &\cong AB \quad \text{(def. of circle centered at } B) \\ AC &\cong BC \quad \text{(common notion 1: things equal to the same thing are equal)} \\ \therefore\ \triangle ABC &\text{ is equilateral.} \end{aligned}

Euclid's pictorial Pythagorean theorem (proof sketch). Draw a square of side $a + b$ two different ways. In each, four copies of the right triangle (legs $a$ , $b$ , hypotenuse $c$ ) are arranged so the leftover region is a square — in the first arrangement two squares of sides $a$ and $b$ , in the second a tilted square of side $c$ . (Showing the tilted region is truly a square uses the fact that the triangle's two acute angles sum to $90^\circ$ .) Equating the two total areas:

\begin{aligned} (a+b)^2 &= 4\cdot \tfrac{1}{2}ab + c^2 \\ a^2 + 2ab + b^2 &= 2ab + c^2 \\ a^2 + b^2 &= c^2. \end{aligned}

Summary

An exterior angle of a triangle is always strictly greater than either of its opposite interior angles (Exterior Angle Theorem).
In any triangle the larger side lies opposite the larger angle, and conversely — an if-and-only-if relationship between side lengths and opposite angles.
The Triangle Inequality $AB + BC > AC$ formalizes "the straight segment is the shortest path"; for arbitrary (possibly collinear) points it becomes $AB + BC \ge AC$ .
The algebraic version $|a + b| \le |a| + |b|$ follows from the piecewise definition of absolute value and the lemmas $x \le |x|$ and $|x| = |{-x}|$ .
Euclid built all of geometry from five postulates and five common notions; each compass-and-straightedge step corresponds exactly to a postulate, so drawing a valid construction is proving the result (e.g. Propositions 1 and 2, isosceles base angles, and a pictorial Pythagorean theorem).
Trisecting an angle, duplicating the cube, and squaring the circle are provably impossible with compass and straightedge — a limitation of Euclid's postulates, not of his logic.