Define triangle congruence through a witness

Try

Place a paper triangle on another triangle by sliding, turning or flipping it. When the entire figures coincide, including their labeled vertices in the required order, the placement gives a congruence witness. For example, the rule (x,y)↦(5−y,2+x) takes A=(0,0), B=(2,0), C=(0,3) to P=(5,2), Q=(5,4), R=(2,2). It is a quarter-turn followed by translation.

Congruence here means the existence of a composition of rigid motions mapping one figure onto the other. We do not define it as “same side lengths,” because proving that side lengths suffice is an important theorem. We also distinguish an unlabeled shape match from the correspondence A↦P, B↦Q, C↦R.

Predict

Predict which corresponding sides must have equal lengths in the example. Does switching Q and R keep the claimed correspondence valid? Next imagine doubling every coordinate of the original triangle. Its angles stay the same, but can a rigid motion map the original onto the enlarged triangle? Decide which preserved quantity would fail before drawing a second diagram.

Experiment

Check the labeled witness by completing the following table. Then inspect one nonvertex point on AB, such as (1,0), and verify that its image lies halfway along PQ.

Source object	Image object	Preserved measurement
AB	PQ	length 2
AC	PR	length 3
BC	QR	length √13
Right angle at A	Right angle at P	90 degrees

If Q and R were switched, AB would need to map to PR, changing length 2 to length 3. The unlabeled target triangle is still congruent, but that proposed labeling is not a valid witness. This distinction prevents a correct theorem from being applied to the wrong side pair.

Understand

Every rigid-motion witness preserves all corresponding side lengths and angles: each elementary motion preserves them, so induction through the finite sequence preserves them at the end. This is the necessary direction of a useful equivalence. The converse says that existing nondegenerate triangles with all corresponding sides and angles equal admit a rigid witness. The next lesson proves SSS from alignment and distance equations; that result supplies the converse using the side equalities alone. Thus the full “sides and angles equal” condition is equivalent to congruence, although its angle information is redundant once all three sides agree. The equivalence is a theorem, not an unexplained alternate definition.

A systematic alignment reduces the construction problem. Translate the first vertex to the origin. Rotate so the second lies on the positive horizontal axis at (c,0), where c>0 is the first side length. If the third vertex lies below that axis, reflect across it. For a nondegenerate triangle the third vertex is not on the axis, so this places it in the upper half-plane. Apply the same normalization to a second triangle. Proving their normalized third vertices coincide will prove congruence by undoing the second triangle’s normalization after the first.

Why do matching vertices under a rigid map match the entire triangle? The basic motion rules are affine: they preserve points of the form (1−t)A+tB for 0≤t≤1. Thus each complete edge maps onto its corresponding edge. Interior points have the form αA+βB+γC with nonnegative coefficients summing to 1; the same affine rule maps them to αP+βQ+γR. Inverse motions give the reverse inclusion. This coordinate explanation supplies a precise version of “the whole paper fits,” without assuming an arbitrary function matching three points must do the same.

Equal angles alone are insufficient. A triangle with sides 3,4,5 and its double with sides 6,8,10 have equal angles, but no rigid witness can double a positive distance. Such a transformation is a similarity, studied later, and must not be mistaken for congruence.

Apply

Explain why a triangle with three collinear vertices is excluded from the normalization argument and later criteria. Its enclosed region has collapsed and its upper-half-plane third vertex does not exist. A zero-length selected side also fails to define a direction for the rotation. State these conditions rather than hiding them behind an apparently ordinary sketch.

In your written witness, list the correspondence first, then the operations, then the preserved quantities. A correct calculation using a different correspondence does not answer the stated problem.

A labeled witness and a scale counterexample

Consider A=(0,0), B=(4,0), C=(1,2) and D=(1,−2). To map ABC to ABD with A and B fixed, reflect in the x-axis. The explicit rule R(x,y)=(x,−y) sends all three named vertices correctly. For any two points U=(u₁,u₂), V=(v₁,v₂), the squared distance after reflection is (u₁−v₁)²+(−u₂+v₂)², equal to the original squared distance. This verifies a rigid motion generally, not just for three sampled sides.

Now replace C by (1,h) and D by (1,−h). The experiment displays squared side lengths so radicals and decimal rounding cannot hide equality. At h=0 all three vertices are collinear. Reflection remains a valid isometry, but a theorem whose hypotheses require a nondegenerate triangle cannot be applied to that collapsed case. Record the distinction explicitly.

A different error is to infer congruence from equal angles alone. Triangles with sides 2,3,4 and 4,6,8 are similar with scale factor two, but corresponding lengths are unequal. A dilation is a similarity witness, not a rigid-motion witness. Neither a convincing picture nor a common angle pattern overrides this invariant.

Before using the retry, draw the two labeled triangles, write a proposed transformation and check each vertex. The practice asks you to diagnose an incorrect map as well as select a correct one. Retain your original drawing and correction for review: a selected answer cannot establish that you can construct an unfamiliar witness.

Construct a labeled rigid-motion witness

Review criterion: Name the vertex correspondence and a distance-preserving map; distinguish congruence from similarity.

Predict, then change one input

Vertical coordinate of C:

2

AC² for A=(0,0), C=(1,y)

5

AD² after reflection D=(1,−y)

5

BC² for B=(4,0)

13

BD² after reflection

13

The reflection fixes A and B and preserves the displayed squared lengths. Prove preservation for arbitrary pairs, not merely these samples.

Explain the changed case: Reflection rather than relabeling

A=(0,0), B=(4,0), C=(1,2), and D=(1,−2). Which map sends labeled triangle ABC to ABD while fixing A and B?

• AC=AD=√5
• BC=BD=√13

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

Dilation by factor 2 about A.A half-turn about the origin.Reflection (x,y) ↦ (x,−y).

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

Derive SSS, SAS and ASA next. Keep track of which facts are inputs, which are consequences of the motions, and where existence of the original nondegenerate triangles is used. That separation prevents a proof from assuming its own conclusion.