Derive the three triangle criteria

Try

Two sticks of known lengths can open through many angles. Adding their included angle fixes a triangle; giving a different angle may not. Congruence criteria explain precisely which measurements determine an existing nondegenerate triangle up to rigid motion. They are not three-letter passwords: each has a construction and a uniqueness argument.

Start with a separate numerical SSS example. Align A=(0,0), B=(10,0), and require AC=√41, BC=√61. For C=(u,v), the squared distances give u²+v²=41 and (u−10)²+v²=61. Subtracting the first equation from the second gives −20u+100=20, hence u=4. Then v²=25, so C=(4,5) or (4,−5). These are reflected placements of the same triangle.

Predict

Predict why subtracting the distance equations removes the vertical coordinate. Does the remaining sign choice create two noncongruent triangles? Next predict why a side and two endpoint angles should determine the third vertex: draw the two rays that it must lie on. Keep the claim “at most one intersection” separate from “an intersection exists.”

Experiment

Draw the worked SSS placements on squared paper with identical scales. Then record the information used at each step rather than adding unproved angle equalities.

Step	Information used	Consequence
Align AB	Equal positive side lengths and motions	Common endpoints (0,0),(10,0)
Subtract squared distances	Other two side lengths	u=4
Recover height	u²+v²=41	v=5 or −5
Reflect if necessary	Reflection is rigid	Third vertices coincide

For SAS, draw a side from (0,0) to (4,0), a ray making 60 degrees above it at the origin, and mark a point distance 3 along that ray. Move the marked point: preserving both its ray and its distance leaves no freedom. For ASA, prescribe one upward ray at each endpoint of a fixed side. Move their intersection while retaining both rays: again there is no freedom when the triangle exists.

Understand

For the general SSS proof, use rigid motions to align a common side at A=(0,0), B=(c,0), c>0. Write AC=b and BC=a. Any third vertex (u,v) obeys u²+v²=b² and (u−c)²+v²=a². Subtraction yields u=(b²+c²−a²)/(2c). The first equation then fixes v². For existing nondegenerate triangles v²>0; the two possible signs are exchanged by reflection across AB. Therefore an allowed composition aligns the entire triangle. We used coordinate distance from an independent Pythagorean dissection theorem, not SSS or a cosine-law formula assumed without explanation. The later altitude-similarity proof of Pythagoras is a second derivation and is not used here.

For SAS, align the first of the two equal sides, and reflect if needed so both third vertices lie on the same side of its line. The equal included angle at the common endpoint determines the same ray for the second side. Its equal positive length selects exactly one point on that ray. Hence the third vertices coincide. The angle must be between the two specified sides; an angle elsewhere does not select this ray with this given distance.

For ASA, align the equal side lying between the two matched angles. On the selected half-plane, each endpoint angle determines a ray. Both third vertices belong to their intersection, which has at most one point because the lines are distinct. Existence comes from the two given triangles, not from a claim that arbitrary angle data always meet. Degenerate or inconsistent data are outside the statement.

These arguments supply sufficient conditions. Their necessary direction is already known: a rigid witness preserves every side and angle. In particular, all corresponding sides and angles are equal if and only if the triangles are congruent, with SSS providing the converse.

Apply

Diagnose SSA by fixing A=(0,0), C=(3,2), and letting B=(t,0) lie on the positive horizontal ray. If BC=√5, then (t−3)²+4=5 gives t=2 or 4. Both have AC=√13, the same BC and the same angle at A, yet different AB lengths. The known angle is not included between the two specified sides. This counterexample refutes general SSA; it does not refute SAS or claim every SSA dataset has two solutions.

For each criterion in practice, write an alignment step, a uniqueness step and the needed nondegeneracy condition. A sketch can guide the proof, but the conditions must survive a differently oriented drawing.

Continue

Continue to parallel-line angle proofs, then dilations and AA similarity. The order matters: similarity will use angle preservation and ray uniqueness, and later quadrilateral theorems will use the congruence criteria just derived. Keep the proof dependencies visible when reusing a result.