The Timpson Lyceum · Geometry (2024)

Congruence of Triangles and Angle Inequalities – Lecture 8 (Geometry)

Overview

This lecture closes out the study of angle congruence by proving the last of the triangle-congruence theorems and then opening the topic of angle inequality. We first complete the trio of congruence theorems by proving Side-Side-Side (SSS), then show that right angles actually exist (rather than being merely defined). From there we build a careful, purely geometric definition of one angle being greater than or less than another — with no notion of degree measure yet — and prove the hard theorem that "$\angle B$ is less than $\angle A$" really does mean the same thing as "$\angle A$ is greater than $\angle B$." We finish with the trichotomy theorem for angles, basic ordering facts, and the result that all right angles are congruent to each other. These tools are foundational: triangle congruence and angle ordering are used constantly throughout the rest of geometry.

Key Concepts

The postulate toolkit (recap). Everything in this chapter rests on four postulates:

• Angle-construction: Given a non-straight angle and a ray on a line, there is one and only one ray on each side of that line forming an angle congruent to the given angle.
• Equivalence relation: Angle congruence is reflexive ($\angle A \cong \angle A$), symmetric ($\angle A \cong \angle B \Rightarrow \angle B \cong \angle A$), and transitive ($\angle A \cong \angle B$ and $\angle B \cong \angle C \Rightarrow \angle A \cong \angle C$).
• Addition/subtraction of angles: Equals added to equals are equal, and equals subtracted from equals are equal. If a whole angle splits into two parts, congruence of the whole and one part forces congruence of the other part.
• Side-Angle-Side (SAS) postulate: If two triangles share a congruent side, an included angle, and a side (in that order), then the two remaining corresponding angles are congruent.

Triangle congruence — notation matters. Writing $\triangle ABC \cong \triangle DEF$ asserts six facts in a fixed correspondence: $\overline{AB}\cong\overline{DE}$, $\overline{BC}\cong\overline{EF}$, $\overline{AC}\cong\overline{DF}$ and $\angle A\cong\angle D$, $\angle B\cong\angle E$, $\angle C\cong\angle F$. Reordering the letters changes the claim and can make it false.

Theorems proved previously (used here as tools).

• Congruent angles have congruent supplements; all straight angles are congruent.
• Vertical angles are congruent.
• Isosceles base-angles theorem: if two sides of a triangle are congruent, the base angles opposite them are congruent (proved by relabeling the triangle two ways and applying SAS).
• The converse: if two angles of a triangle are congruent, the opposite sides are congruent.
• SAS theorem (stronger than the postulate) and ASA theorem give full triangle congruence.

Side-Side-Side (SSS) congruency theorem. If two triangles have all three pairs of corresponding sides congruent, the triangles are congruent. This is the third and final congruence theorem of the chapter.

Existence of right angles. A right angle is defined as an angle congruent to its own supplement. A definition does not guarantee existence (you can name an angle "greater than a straight angle," but no such angle exists, since a straight angle is the largest). The lecture proves that a right angle genuinely exists by constructing one.

Comparing angles (no degrees yet). An angle is purely two rays from a point — there is no number attached. To compare $\angle A$ and $\angle B$, construct $\angle B'$ congruent to $\angle B$ sharing one side with $\angle A$, with its other ray on the same side as $\angle A$'s other side. Then:

• If $\angle B'$'s free ray lands in the interior of $\angle A$, we say $\angle B < \angle A$.
• If it lands in the exterior of $\angle A$, we say $\angle B > \angle A$.
• A straight angle is defined to be larger than every non-straight angle (orientation/"going around the circle" is meaningless without measure).

The hard ordering theorem. "$\angle B < \angle A$" and "$\angle A > \angle B$" are defined by constructions on different angles, so they are not automatically the same statement. The chapter's hardest theorem proves they are equivalent.

Trichotomy (Theorem 6.9). For any two angles, exactly one holds: $\angle A > \angle B$, $\angle A \cong \angle B$, or $\angle A < \angle B$. (When the construction ray lands in the exterior, on the side, or in the interior, respectively.)

Ordering facts (Theorem 6.10). Congruence and inequality combine as expected, e.g. $\angle A \cong \angle B$ and $\angle B < \angle C \Rightarrow \angle A < \angle C$; inequalities are transitive.

Obtuse / acute. An angle greater than a right angle is obtuse; one less than a right angle is acute.

All right angles are congruent. Any two right angles are congruent — proved without measuring, by contradiction.

Perpendicular lines. Two lines $\ell$ and $m$ are perpendicular ($\ell \perp m$) when their angles of intersection are right angles.

Worked Example

Proof that all right angles are congruent (by contradiction).

Let $\angle A$ and $\angle B$ be right angles with supplements $\angle A'$ and $\angle B'$, so by definition $\angle A \cong \angle A'$ and $\angle B \cong \angle B'$. Suppose, for contradiction, $\angle A \not\cong \angle B$. By trichotomy one is larger; without loss of generality take $\angle A < \angle B$.

Since $\angle A < \angle B$, construct a ray interior to $\angle B$ giving $\angle C \cong \angle A$, and let $\angle C'$ be the supplement of $\angle C$ along the shared line. Because supplements of congruent angles are congruent, $\angle A' \cong \angle C'$. Now chain the congruences, using that $\angle A$ is a right angle:

The middle line uses that $\angle C$'s ray is interior to $\angle B$ (so $\angle C'$'s ray is exterior to $\angle B'$, making $\angle C' > \angle B'$). Substituting $\angle C' \cong \angle B$ on the left of $\angle C' > \angle B'$ and $\angle B' \cong \angle B$ on the right gives

$\angle B > \angle B,$

which is impossible — no angle exceeds itself. The contradiction forces $\angle A \cong \angle B$. $\blacksquare$

Sketch — why SSS holds. Given $\triangle ABC$ and $\triangle A'B'C'$ with all three sides congruent, construct on the far side of $\overline{A'C'}$ a triangle $\triangle A'C'E'$ congruent to $\triangle ABC$ by SAS (copy $\angle A$ and mark off the matching side length). Dropping the segment $\overline{B'E'}$ creates two isosceles triangles whose base angles are congruent; adding these equal base angles ($\angle A'B'C' \cong \angle A'E'C'$) gives a SAS match between $\triangle A'B'C'$ and the constructed $\triangle A'E'C'$. Hence $\triangle A'B'C' \cong \triangle A'E'C' \cong \triangle ABC$ by transitivity.

Summary

• SSS is the third triangle-congruence theorem: three pairs of congruent sides force the triangles congruent. Its proof copies one triangle by SAS, then uses isosceles base angles plus angle addition to match the original.
• Right angles exist — proved by construction (with three position cases for the chosen point), not assumed; a right angle is one congruent to its supplement.
• Angle order is purely geometric: construct a congruent copy sharing a side and see whether its free ray falls interior (smaller), on the side (congruent), or exterior (larger). No degree measure is available yet.
• The chapter's hardest theorem shows $\angle B < \angle A$ is equivalent to $\angle A > \angle B$; trichotomy then guarantees exactly one of $<$, $\cong$, $>$ for any two angles.
• All right angles are congruent, proved by contradiction using trichotomy and supplements — leading to the absurd $\angle B > \angle B$.
• Two lines are perpendicular when they meet in right angles; angles larger/smaller than a right angle are obtuse/acute.