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Congruency with Mathematica

Open the video directly or download the edited lecture transcript.

Accessibility note: the edited transcript supplies a complete text route through the mathematical ideas while synchronized captions are not available.

Central question

Which transformations of \(\mathbb R^2\) preserve every distance, and how do those transformations explain the classical triangle-congruence criteria?

Learning outcomes

By the end, you should be able to:

compute with rotation, reflection, and translation maps;
distinguish \(O(2)\), \(SO(2)\), \(E(2)\), and \(SE(2)\);
prove that \(x\mapsto Qx+b\) preserves distance when \(Q\) is orthogonal;
classify every distance-preserving map of the plane;
prove SSS, SAS, and ASA from Euclidean motions;
verify the invariants with JSXGraph and Mathematica.

Prerequisites

vectors and \(2\times2\) matrix multiplication;
transpose, determinant, and inverse;
dot product and Euclidean norm;
elementary facts about circles, rays, and angles;
triangle angle sum and the Pythagorean theorem.

Precision that matters

A reflection reverses orientation, but it still preserves every distance. Therefore ordinary congruence is governed by all of \(E(2)\). If reflections are forbidden, one obtains direct congruence, governed by the subgroup \(SE(2)\).

Two matrix families

For \(t\in\mathbb R\), define

\[ R(t)= \begin{pmatrix}\cos t&-\sin t\\ \sin t&\cos t\end{pmatrix}, \qquad M(t)= \begin{pmatrix}\cos t&\sin t\\ \sin t&-\cos t\end{pmatrix}. \]

\(R(t)\) rotates counterclockwise through angle \(t\). The matrix \(M(t)\) reflects across the line through the origin whose direction angle is \(t/2\). Both satisfy \(Q^{\mathsf T}Q=I\), but \(\det R(t)=1\) and \(\det M(t)=-1\).

Matrix products

\[ \begin{aligned} R(t)R(s)&=R(t+s),& R(t)M(s)&=M(t+s),\\ M(t)R(s)&=M(t-s),& M(t)M(s)&=R(t-s). \end{aligned} \]

In particular, \(R(t)^{-1}=R(-t)\), \(M(t)^{-1}=M(t)\), and the identity is \(R(0)=I\).

Exact vector images

For \(w=(\tfrac12,\tfrac12)\),

\[ R\!\left(\frac{\pi}{3}\right)w =\left(\frac{1-\sqrt3}{4},\frac{1+\sqrt3}{4}\right), \] \[ M\!\left(\frac{\pi}{3}\right)w =\left(\frac{1+\sqrt3}{4},\frac{\sqrt3-1}{4}\right). \]

The orthogonal groups

\[ O(2)=\{Q\in M_2(\mathbb R):Q^{\mathsf T}Q=I\},\qquad SO(2)=\{Q\in O(2):\det Q=1\}. \]

Every matrix in \(O(2)\) has determinant \(1\) or \(-1\). The determinant-\(1\) part \(SO(2)\) consists exactly of the rotations \(R(t)\).

Theorem: orthogonal matrices preserve length and distance

If \(Q^{\mathsf T}Q=I\), then for every \(z\in\mathbb R^2\),

\[ \|Qz\|^2=(Qz)^{\mathsf T}(Qz) =z^{\mathsf T}Q^{\mathsf T}Qz =z^{\mathsf T}z=\|z\|^2. \]

Norms are nonnegative, so \(\|Qz\|=\|z\|\). Substituting \(z=x-y\) gives \(\|Qx-Qy\|=\|Q(x-y)\|=\|x-y\|\). \(\square\)

Translations and Euclidean motions

A translation by \(b\in\mathbb R^2\) is \(T_b(x)=x+b\). For example, \[ \left(\frac13,\frac23\right)+\left(\frac23,\frac12\right) =\left(1,\frac76\right). \] A general Euclidean motion has the form \[ F_{Q,b}(x)=Qx+b,\qquad Q\in O(2),\ b\in\mathbb R^2. \]

\[ E(2)=\mathbb R^2\rtimes O(2),\qquad SE(2)=\mathbb R^2\rtimes SO(2). \]

Composition and inversion are \[ (Q,b)(P,c)=(QP,Qc+b),\qquad (Q,b)^{-1}=(Q^{\mathsf T},-Q^{\mathsf T}b). \]

Interactive laboratory 1

Move one triangle by \(F(x)=Qx+b\)

Choose a rotation or reflection, then vary \(t\) and the translation \(b=(u,v)\). The red triangle is fixed; the gold triangle is its image. Dashed segments join corresponding vertices.

Orthogonal part

Parameter \(t\) 60°

Horizontal shift \(u\) 1.00

Vertical shift \(v\) 0.50

The parameter in \(M(t)\) is twice the angle of the reflecting line. The signed area changes sign exactly when \(\det Q=-1\).

Rotation \(R(60^\circ)\), followed by translation \((1,0.5)\).

Invariant comparison: original and image triangle
Quantity	Original	Image	Absolute error
Side AB\(AB\)	1.000	1.000	0.000e+0
Side AC\(AC\)	0.707	0.707	0.000e+0
Side BC\(BC\)	0.707	0.707	0.000e+0
Signed area	+0.250	+0.250	same orientation

Text alternative: orthogonal maps preserve the three side lengths. A rotation preserves the sign of the oriented area; a reflection reverses it.

Ordinary congruence

Subsets \(A,B\subseteq\mathbb R^2\) are congruent when there exists \(F\in E(2)\) such that \(F(A)=B\). For ordered triangles, the chosen isometry must respect the stated vertex correspondence.

Direct congruence

\(A\) and \(B\) are directly congruent when the motion can be chosen in \(SE(2)\). Direct congruence permits rotations and translations but not reflections.

Why congruence is an equivalence relation

Reflexive: the identity \(F_{I,0}\) maps every figure to itself.
Symmetric: if \(F(A)=B\), then the Euclidean motion \(F^{-1}\) maps \(B\) to \(A\).
Transitive: if \(F(A)=B\) and \(G(B)=C\), then the Euclidean motion \(G\circ F\) maps \(A\) to \(C\).

Classification theorem for plane isometries

A map \(F:\mathbb R^2\to\mathbb R^2\) preserves every Euclidean distance if and only if there are \(Q\in O(2)\) and \(b\in\mathbb R^2\) for which \(F(x)=Qx+b\).

Proof of the classification theorem

The forward formula has already been checked: \[ \|F(x)-F(y)\|=\|Q(x-y)\|=\|x-y\|. \]

Conversely, suppose \(F\) preserves distances. Put \(b=F(0)\) and \(G(x)=F(x)-b\). Then \(G(0)=0\), \(G\) preserves distance, and \(\|G(x)\|=\|x\|\). The polarization identity gives \[ \langle G(x),G(y)\rangle =\frac{\|G(x)\|^2+\|G(y)\|^2-\|G(x)-G(y)\|^2}{2} =\langle x,y\rangle . \]

Thus \(G(e_1)\) and \(G(e_2)\) form an orthonormal basis. Let \(Q\) have those two vectors as its columns, so \(Q^{\mathsf T}Q=I\). For \(x=x_1e_1+x_2e_2\), \[ \langle G(x),G(e_i)\rangle=\langle x,e_i\rangle=x_i. \] Coordinates in the orthonormal basis \(\{G(e_1),G(e_2)\}\) therefore give \(G(x)=x_1G(e_1)+x_2G(e_2)=Qx\). Hence \(F(x)=Qx+b\). \(\square\)

Vertex order is mathematical data

Writing \(\triangle ABC\cong\triangle XYZ\) asserts \(A\leftrightarrow X\), \(B\leftrightarrow Y\), and \(C\leftrightarrow Z\). Consequently \(AB=XY\), \(BC=YZ\), \(CA=ZX\), with the corresponding angle equalities.

SSS congruence

If \(AB=XY\), \(BC=YZ\), and \(CA=ZX\), then \(\triangle ABC\cong\triangle XYZ\).

Proof by aligning the common base

First choose a Euclidean motion sending \(A\) to \(X\) and \(B\) to \(Y\); this is possible because \(AB=XY\). After this alignment, the image of \(C\) must lie on the circle centered at \(X\) of radius \(AC=XZ\), and on the circle centered at \(Y\) of radius \(BC=YZ\). Two such circles have at most two intersection points, which are reflections of one another across line \(XY\). The point \(Z\) is one of them. If needed, compose with that reflection, which fixes \(X\) and \(Y\), to send the image of \(C\) to \(Z\). The resulting Euclidean motion maps all three vertices correctly. \(\square\)

SAS congruence

If \(AB=XY\), \(AC=XZ\), and the included angles \(\angle BAC=\angle YXZ\), then \(\triangle ABC\cong\triangle XYZ\).

Proof by a ray–circle intersection

Align \(A\) with \(X\) and \(B\) with \(Y\). If necessary, reflect across line \(XY\) so the image of \(C\) lies on the same side as \(Z\). The included angle fixes the ray from \(X\), while the equality \(AC=XZ\) fixes the distance from \(X\). Exactly one point on that ray lies at the required distance, so the image of \(C\) is \(Z\). \(\square\)

ASA congruence

If \(AB=XY\), \(\angle A=\angle X\), and \(\angle B=\angle Y\), then \(\triangle ABC\cong\triangle XYZ\).

Proof by intersecting two rays

Align \(A\) with \(X\) and \(B\) with \(Y\), again choosing the correct side of the base by reflection if needed. The angle at \(X\) fixes the ray containing the image of \(C\); the angle at \(Y\) fixes a second ray. Nondegeneracy ensures those rays meet in exactly one point. Since \(Z\) satisfies both angle conditions, that intersection is \(Z\). \(\square\)

Useful consequences

AAS: the triangle angle sum supplies the third angle, reducing to ASA.
HL: in right triangles, the Pythagorean theorem recovers the missing leg, reducing to SSS.

Data that do not suffice

AAA fixes shape but not scale.
SSA may produce two noncongruent triangles—the ambiguous case.

Interactive laboratory 2

See where the third vertex can live

Switch among the criteria. Each diagram turns the proof into an intersection problem: circle–circle for SSS, ray–circle for SAS, and ray–ray for ASA.

In the SSS view, the two possible third vertices are mirror images across the base. They are different placements but belong to the same congruence class.

SSS: two circle intersections are exchanged by reflection across the base.

Construction logic without the graph
Criterion	Given data	Location of \(C\)
SSS	\(AB=4,\ AC=3.2,\ BC=2.6\)	Intersection of two circles; two reflected positions
SAS	\(AB=4,\ AC=3.2,\ \angle A=55^\circ\)	Positive-ray intersection with one circle
ASA	\(AB=4,\ \angle A=55^\circ,\ \angle B=48^\circ\)	Unique intersection of two interior rays

These Wolfram Language cells reproduce the algebra behind the lesson. Run each block in a fresh notebook; the assertions return exact symbolic results.

Companion Mathematica notebook

Open the instructor-supplied notebook to run, modify, and extend the computations presented in this section.

OpenCongruency.nb (opens in a new tab)

1. Define and verify \(R(t)\) and \(M(t)\)

ClearAll[R, M, t, s];
R[t_] := {{Cos[t], -Sin[t]}, {Sin[t], Cos[t]}};
M[t_] := {{Cos[t],  Sin[t]}, {Sin[t], -Cos[t]}};

FullSimplify[Transpose[R[t]].R[t]]
FullSimplify[Transpose[M[t]].M[t]]
FullSimplify[Det[R[t]]]
FullSimplify[Det[M[t]]]

FullSimplify[R[t].R[s] == R[t + s]]
FullSimplify[R[t].M[s] == M[t + s]]
FullSimplify[M[t].R[s] == M[t - s]]
FullSimplify[M[t].M[s] == R[t - s]]

2. Apply an affine Euclidean motion

ClearAll[F, Q, b, x];
Q = R[Pi/3];
b = {2, -1};
F[x_] := Q.x + b;

triangle = {{0, 0}, {1, 0}, {1/2, 1/2}};
image = F /@ triangle;

Graphics[{
  {Directive[Red, Thick], Polygon[triangle]},
  {Directive[RGBColor[0.66, 0.38, 0], Thick], Polygon[image]},
  Black, PointSize[Large], Point /@ Join[triangle, image]
  },
 Axes -> True, PlotRange -> All, ImageSize -> Large]

3. Build the interactive motion

Manipulate[
 Module[{Q, F, tri, img},
  Q = If[motion == "Rotation", R[t], M[t]];
  F[x_] := Q.x + {u, v};
  tri = {{0, 0}, {1, 0}, {1/2, 1/2}};
  img = F /@ tri;
  Graphics[{
    {Red, Thick, Polygon[tri]},
    {Darker[Orange], Thick, Polygon[img]},
    {Purple, Dashed, MapThread[Line[{#1, #2}] &, {tri, img}]}
    }, Axes -> True, PlotRange -> {{-3, 4}, {-3, 3}}],
  {{motion, "Rotation"}, {"Rotation", "Reflection"}},
  {t, -Pi, Pi}, {u, -2, 2}, {v, -2, 2}]

4. Test distance and orientation invariants

ClearAll[pairDistances, signedArea];
pairDistances[pts_] :=
  EuclideanDistance @@@ Subsets[pts, {2}];

signedArea[{a_, b_, c_}] :=
  Det[Transpose[{b - a, c - a}]]/2;

tri = {{0, 0}, {1, 0}, {1/2, 1/2}};
rotated = (# + {1, 1/2}) & /@ (R[Pi/3].# & /@ tri);
reflected = (# + {1, 1/2}) & /@ (M[Pi/3].# & /@ tri);

FullSimplify[pairDistances[tri] == pairDistances[rotated]]
FullSimplify[pairDistances[tri] == pairDistances[reflected]]
{signedArea[tri], signedArea[rotated], signedArea[reflected]}

Example 1: rotate by \(\pi/3\), then translate by \((2,-1)\)

For \(F(x)=R(\pi/3)x+(2,-1)\), \[ F(0,0)=(2,-1),\qquad F(1,0)=\left(\frac52,-1+\frac{\sqrt3}{2}\right), \] \[ F\!\left(\frac12,\frac12\right) =\left(2+\frac{1-\sqrt3}{4}, -1+\frac{1+\sqrt3}{4}\right). \] Since \(\det R(\pi/3)=1\), the image triangle is directly congruent to the original.

Example 2: identify the motion from its formula

Consider \[ F(x,y)=\left(\frac{\sqrt3}{2}x+\frac12y+3,\, \frac12x-\frac{\sqrt3}{2}y-2\right). \] Its linear part is \(M(\pi/6)\), so \(F\in E(2)\) and \(\det Q=-1\). It is a reflection followed by a translation: it gives ordinary congruence, but not direct congruence.

Example 3: recover the inverse motion

If \(F(x)=Qx+b\), then \[ F^{-1}(y)=Q^{\mathsf T}y-Q^{\mathsf T}b. \] Indeed, \(Q^{\mathsf T}(Qx+b)-Q^{\mathsf T}b=x\). This calculation is the symmetry step in the equivalence-relation proof.

Non-example: a dilation is not a congruence

The map \(D(x)=2x\) has matrix \(2I\), and \((2I)^{\mathsf T}(2I)=4I\ne I\). It doubles every distance, so it produces similar figures but generally not congruent figures.

Quick checks

1. Which group governs ordinary congruence in the plane? \(SO(2)\) \(SE(2)\) \(E(2)\)

2. If \(Q\in O(2)\), which identity is always true? \(Q^{-1}=-Q\) \(Q^{-1}=Q^{\mathsf T}\) \(\det Q=1\)

3. Why do the two possible SSS third vertices give congruent triangles? A reflection across the base exchanges them. Their areas need not agree. AAA implies congruence.

Six common errors

excluding reflections from ordinary congruence;
forgetting the translation vector in a general isometry;
confusing \(M(t)\) with reflection across a line at angle \(t\);
matching vertices in the wrong order;
using AAA as a congruence criterion;
treating SSA as unambiguous.

Exit ticket

In two sentences, explain why distance preservation allows a reflection, while orientation preservation excludes it. Name the two relevant groups.

Exercises

Verify directly that \(M(t)^2=I\).
Compute \(R(\pi/4)M(\pi/3)\) and identify the resulting reflection parameter.
For \(F(x)=R(\pi/2)x+(3,-1)\), compute \(F^{-1}\).
Prove that the composition of two Euclidean motions is a Euclidean motion.
Use the classification proof to explain why a plane isometry is determined by the images of three noncollinear points.
Construct numerical SSA data that produce two noncongruent triangles.

The whole lesson in one implication chain

\[ Q^{\mathsf T}Q=I \Longrightarrow \|Qz\|=\|z\| \Longrightarrow F(x)=Qx+b\text{ preserves distance} \Longrightarrow \text{congruent images}. \]

The classification theorem reverses the central implication: every distance-preserving map of the plane has exactly this affine-orthogonal form. SSS, SAS, and ASA then state that limited triangle data are sufficient to produce such a motion.

Ordinary congruence\(E(2)\)

Direct congruence\(SE(2)\)

Invariantdistance

Vocabulary

Orthogonal matrix: A matrix \(Q\) satisfying \(Q^{\mathsf T}Q=I\).
Euclidean motion: A map \(x\mapsto Qx+b\) with \(Q\in O(2)\).
Orientation: The sign of an ordered area; rotations preserve it and reflections reverse it.
Congruence criterion: A data set sufficient to determine one triangle up to Euclidean motion.

References and provenance

Congruency.nb, instructor-supplied Mathematica notebook, exported as Congruency.pdf. Open the notebook (opens in a new tab) .
The instructor’s lecture transcript, edited for mathematical clarity.
JSXGraph documentation for the coordinate constructions.
Wolfram Language documentation for the computation blocks.

▾ 1. Lecture video and transcript