Congruence axioms and SAS

Try

Put A=(0,0), B=(4,0), and require AC=3 with included angle BAC=60°, with C above AB. Then C=(3/2,3√3/2). If the half-plane condition is removed, its reflection below AB is another placement with the same side lengths and unsigned angle.

This coordinate calculation illustrates side–angle–side congruence, but a synthetic development cannot start by assuming all Euclidean coordinate formulas. A Hilbert-style congruence block introduces segment equivalence, unique segment transport on a ray, compatible segment addition, angle transport on a specified side, and a SAS principle connecting these notions.

Predict

Predict whether three matching angles force equal lengths. In the Euclidean plane, scaling a triangle preserves its angles while changing every length. Predict whether equal segments can be transported uniquely along a full line without a direction choice. Usually there are two possible endpoint locations, one on each ray.

Uniqueness statements must specify the available side or ray. Otherwise a reflection supplies a second valid placement rather than a contradiction.

Experiment

Compare coordinate placements with b=3, c=4 and included angle θ.

θ	C above the base	Squared BC
60°	(3/2,3√3/2)	13
90°	(0,3)	25
120°	(−3/2,3√3/2)	37

Compute BC² by subtracting coordinates and squaring. The table changes one angle while holding two lengths fixed. It does not define neutral congruence by a Euclidean cosine formula. Its purpose is to show what the same SAS data determine in the real Euclidean model.

Understand

For a coordinate SAS proof, place the first side on the positive x-axis: A=(0,0), B=(c,0), c>0. If AC=b>0 and the included angle is θ in (0,π), the required half-plane makes C=(b cos θ,b sin θ). The coordinates are unique. A rigid placement of another triangle with identical data therefore makes all three vertices agree. Reflection handles the opposite orientation.

A synthetic example uses SAS without trigonometry. Suppose AB is congruent to AC. Compare triangle ABC with triangle ACB. The two sides at A correspond in reverse order, and their included angle is the same unsigned angle. SAS identifies the triangles, giving equality of the base angles at B and C.

This is a correspondence argument, not a claim that the two triangles occupy different parts of the drawing. Explicit vertex order prevents accidentally equating an angle with an unrelated one.

Apply

Diagnose “SAS follows in neutral geometry from c²=a²+b²−2ab cos C.” That formula is a Euclidean metric theorem. Using it before establishing the relevant Euclidean structure can smuggle in the parallel postulate. In a synthetic neutral theory, SAS is part of the congruence axioms or is derived from an explicitly different axiom system.

For optional depth, distinguish congruence as a primitive relation from congruence defined by an isometry. Proving those descriptions coincide requires a representation or motion theorem. Our coordinate lessons will establish the isometry description inside R²; they do not silently prove a categoricity theorem for every synthetic system.

Continue

Record which facts are assumptions and which are conclusions in each congruence proof. Continue to continuity and coordinate fields to see why transporting or intersecting geometric objects can require more than rational arithmetic.