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The Timpson Lyceum · Geometry (2024)

Congruence of Angles and Triangles – Lecture 7 (Geometry)

Congruence of Angles and Triangles – Lecture 7

Overview

This lecture extends the idea of congruence from segments (developed earlier) to angles, and then to whole triangles. Every congruence statement in geometry ultimately reduces to segments being congruent or angles being congruent; a triangle congruence is just a claim that all corresponding sides and angles match. We restate the angle versions of the construction, equivalence, and addition/subtraction postulates, introduce the Side-Angle-Side (SAS) postulate, and use these tools to prove a chain of results: congruent angles have congruent supplements, vertical angles are congruent, the base angles of an isosceles triangle are congruent (and its converse), and the SAS and ASA triangle congruence theorems.

Key Concepts

Congruence reduces to segments and angles. Two polygons are congruent exactly when all corresponding sides and all corresponding angles are congruent. The same symbol ≅\cong is used for both segment and angle congruence; context determines which is meant.

Angle-construction postulate. Given a non-straight angle AA and a ray r′r' from a point A′A' on a line L′L', there exists a unique ray s′s' on each side of L′L' such that the angle formed equals angle AA. (This mirrors the segment-copying postulate: one and only one congruent copy on each side.)

Congruence is an equivalence relation. For both segments and angles:

Angle addition and subtraction. With adjacent angles set up so the whole is composed of the non-shared sides:

Straight angles. All straight angles are congruent to one another (taken as a postulate alongside "all right angles are congruent").

Supplements. Two angles are supplementary if their non-shared sides form a straight line (they "add to a straight angle"). Right angle: an angle that is congruent to its own supplement.

Side-Angle-Side postulate (Postulate 54). If two triangles △ABC\triangle ABC and △A′B′C′\triangle A'B'C' have AB‾≅A′B′‾\overline{AB}\cong\overline{A'B'}, ∠A≅∠A′\angle A \cong \angle A', and AC‾≅A′C′‾\overline{AC}\cong\overline{A'C'} — that is, two sides and the included angle (the angle between them) — then the remaining angles match: ∠B≅∠B′\angle B \cong \angle B' and ∠C≅∠C′\angle C \cong \angle C'. The included-angle order matters; an angle not between the two sides does not suffice. The postulate alone gives only the two angles; it does not by itself assert the third side is congruent.

Isosceles triangle. A triangle with two congruent sides. The angles opposite those sides are the base angles. An equilateral triangle is a special case of an isosceles triangle.

Triangle congruence. △ABC≅△EFG\triangle ABC \cong \triangle EFG means corresponding parts match in the listed order: AB‾≅EF‾\overline{AB}\cong\overline{EF}, BC‾≅FG‾\overline{BC}\cong\overline{FG}, AC‾≅EG‾\overline{AC}\cong\overline{EG} and ∠A≅∠E\angle A\cong\angle E, ∠B≅∠F\angle B\cong\angle F, ∠C≅∠G\angle C\cong\angle G. Equality of triangles (==) is stronger still: it means the labeled vertices are literally the same points, so △A′C′B′=△ABC\triangle A'C'B' = \triangle ABC identifies A′ ⁣= ⁣AA'\!=\!A, C′ ⁣= ⁣BC'\!=\!B, B′ ⁣= ⁣CB'\!=\!C.

Midpoint. MM is the midpoint of AB‾\overline{AB} iff MM is between AA and BB and AM‾≅MB‾\overline{AM}\cong\overline{MB}.

Theorems proved this lecture.

Worked Example

Claim: If two angles are congruent, then their supplements are congruent.

Let ∠B\angle B and ∠B′\angle B' be the supplements of ∠A\angle A and ∠A′\angle A' along lines LL and L′L', with ∠A≅∠A′\angle A \cong \angle A'. Since all straight angles are congruent, ∠L≅∠L′\angle L \cong \angle L'. The supplement is what remains after removing the angle from the straight angle, so we subtract congruent pieces from congruent wholes:

whole−part=supplement:∠L−∠A=∠B,∠L′−∠A′=∠B′∠L≅∠L′ and ∠A≅∠A′ ⟹ ∠B≅∠B′.\begin{aligned} \text{whole}-\text{part}=\text{supplement}:\quad &\angle L - \angle A = \angle B,\qquad \angle L' - \angle A' = \angle B' \\ \angle L \cong \angle L' \ \text{and}\ \angle A \cong \angle A' \ \Longrightarrow\ &\angle B \cong \angle B'. \end{aligned}

This uses equals subtracted from equals are equal. The same theorem immediately yields vertical angles are congruent: if ∠A\angle A and ∠A′\angle A' are vertical angles and ∠B\angle B is an adjacent angle, then ∠A\angle A and ∠A′\angle A' are both supplements of ∠B\angle B; since ∠B≅∠B\angle B \cong \angle B (reflexive), their supplements ∠A\angle A and ∠A′\angle A' are congruent.

Summary