SAS congruence — SAS congruence · Principia

Dependencies: SAS similarity axiom.

Statement

Let $\triangle ABC$ and $\triangle A'B'C'$ both be non-degenerate (i.e. their three vertices are not collinear), and suppose

|AB| = |A'B'|,\qquad |AC| = |A'C'|,\qquad \angle BAC = \angle B'A'C'.

Then the two triangles are congruent:

\triangle ABC \;\cong\; \triangle A'B'C',

i.e. the three pairs of corresponding sides are equal, and the three pairs of corresponding angles are equal.

$SAS congruence: |AB|=|A'B'|, |AC|=|A'C'|, \angle BAC = \angle B'A'C' ⇒ \triangle ABC \cong \triangle A'B'C'$

Proof

SAS congruence is the degenerate version of axiom IV — the special case where the similarity ratio $k$ is set to $1$ .

Axiom IV (SAS similarity) states: if

\frac{|A'B'|}{|AB|} \;=\; \frac{|A'C'|}{|AC|} \;=\; k,\qquad \angle BAC \;=\; \angle B'A'C',

then $\triangle ABC \sim \triangle A'B'C'$ , and furthermore

\frac{|B'C'|}{|BC|} = k,\quad \angle ABC = \angle A'B'C',\quad \angle ACB = \angle A'C'B'.

Substituting the hypotheses $|AB|=|A'B'|$ and $|AC|=|A'C'|$ immediately gives

k \;=\; \frac{|A'B'|}{|AB|} \;=\; \frac{|AB|}{|AB|} \;=\; 1.

Step 1: invoking axiom IV with k = |A'B'|/|AB| = |A'C'|/|AC|; the two pairs of equal sides in the hypothesis pin k to 1

Substituting $k=1$ back into the three conclusions of similarity:

$\dfrac{|B'C'|}{|BC|} = 1$ , so $|B'C'| = |BC|$ ;
$\angle ABC = \angle A'B'C'$ (independent of $k$ ; always holds for similarity);
$\angle ACB = \angle A'C'B'$ (likewise).

Adding the directly-given hypotheses $|AB|=|A'B'|$ , $|AC|=|A'C'|$ , $\angle BAC = \angle B'A'C'$ , all three pairs of corresponding sides and all three pairs of corresponding angles are equal, so by definition the two triangles are congruent. $\qquad\blacksquare$

$Step 2: k=1 pins down the remaining three conclusions of similarity, |B'C'|=|BC|, \angle B = \angle B', \angle C = \angle C'; with three pairs of equal sides + three pairs of equal angles ⇒ ≅$

Immediate consequences

"Congruence" = "similarity with $k=1$ ": SAS congruence formally absorbs "congruence" into the framework of "similarity" as a special case. From here on, every argument that needs to "copy a triangle" (Base angles of an isoceles triangle are equal for isosceles base angles, SSS congruence for SSS, Perpendicular bisector ⇔ equidistant from endpoints for the perpendicular-bisector locus, …) can directly invoke SAS congruence without going back to axiom IV.
Unified finish for combined tests: ASA (ASA congruence), SSS (SSS congruence), AAS (AAS congruence), HL (HL (hypotenuse-leg) congruence), and so on, all finish by translating their information into "two sides plus an included angle" and applying SAS congruence once more.
Algebraic version of "rigid motion": translating Euclid's "two triangles can be made to coincide" into "corresponding sides and corresponding angles are equal item by item" — SAS congruence turns Euclid's "superposition" gesture into a purely algebraic system of equalities, eliminating the need for geometric intuition about rigid motions.

Remarks

SAS congruence is the cheapest theorem in the four-axiom system: the proof is merely setting the parameter $k$ in axiom IV to $1$ . Yet in middle-school geometry it is the most heavily used — most subsequent proofs treat it as a "basic move".

The brevity of SAS congruence is precisely an advantage of the Birkhoff system over Hilbert's: Hilbert posits SAS congruence as a separate axiom (Group III), and treats "similarity" as a separate downstream theory, whereas Birkhoff merges the two into axiom IV, with SAS congruence simply being a degenerate invocation.