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Subsection 2.3.5 Rotation and Congruence

Learner Experience 2.3.45.

Work in groups
Copy the trangle \(ABC\) and the point \(D\) on a graph paper. Using a ruler and a protractor, rotate the triangle \(-90^\circ\) about point \(D.\)
Draw a dotted line to connect vertex \(A\) to point \(D\)
Place a protractor at the line \(AD\) with the centre of the protractor at \(D\) and measure \(90^\circ.\) Using a ruler draw \(DA'\) such that \(AD\,=\,DA'.\)
Repeat the step above for for vertices \(B \text{ and} C.\)

Exploration 2.3.46. Rotation and Congruence.

Instructions.
Use this interactive board to explore rotations around point \(D\text{:}\)
  • Adjust the Angle: Drag the slider at the top left to change the rotation angle. Notice that positive angles rotate the triangle anticlockwise, while negative angles rotate it clockwise.
  • Move the Centre: Drag the red point \(D\) to see how changing the centre of rotation affects the final position of the image.
  • Observe Congruence: No matter the angle or centre of rotation, notice that the size and shape of the green triangle always exactly match the blue triangle.
Figure 2.3.47. Interactive Activity: Rotation of Positive and Negative Angles

Key Takeaway 2.3.48. Rotation produces congruent figures.

Congruence is a relationship between two figures that are identical in size and shape. Rotation is a transformation that repositions a figure while preserving its shape and size. Because rotation preserves every distance and every angle, a figure and its image under rotation are always congruent.
In the animation below, \(\Delta ABC\) (blue) is fixed and \(\Delta A'B'C'\) (red) rotates about the centre \(D\text{.}\) However far the image turns, each side keeps the length of the side it came from, so the two triangles remain congruent throughout the motion.
Figure 2.3.49. As \(\Delta A'B'C'\) rotates about \(D\text{,}\) corresponding side lengths stay equal, so the image is always congruent to the object.
Because \(\Delta ABC\) and \(\Delta A'B'C'\) have the same shape and size, they are said to be directly congruent. Note in particular that:
  • Rotation is a rigid transformation — it preserves distances and angles, so the image is always congruent to the object.
  • The corresponding sides of the object and image are equal in length, and every corresponding interior angle is unchanged.
  • Direct congruence: rotation preserves orientation, unlike reflection, which reverses it.
  • Proving congruence after rotation: calculate the side lengths with the distance formula, check that corresponding sides are equal, then conclude congruence by SSS (or SAS).
  • Rotation about any point preserves congruence, not just rotation about the origin.

Example 2.3.50.

Triangle \(ABC\) has vertices \(A(2,1)\text{,}\) \(B(5,1)\) and \(C(2,5)\text{.}\) The triangle is rotated \(90^\circ\) anticlockwise about the origin \(O(0,0)\text{.}\)
Find the image \(A'B'C'\) and show that \(\Delta ABC\) and \(\Delta A'B'C'\) are congruent.
Solution.
Under a \(90^\circ\) anticlockwise rotation about the origin, the point \((x,y)\) maps to \((-y,x)\text{.}\) Applying this rule:
\begin{equation*} A(2,1)\mapsto A'(-1,2)\text{,} \end{equation*}
\begin{equation*} \qquad B(5,1)\mapsto B'(-1,5)\text{,} \end{equation*}
\begin{equation*} \qquad C(2,5)\mapsto C'(-5,2). \end{equation*}
Compare corresponding side lengths using the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\text{:}\)
\begin{align*} AB \amp = \sqrt{(5-2)^2+(1-1)^2} = 3\\ A'B' \amp = \sqrt{(-1+1)^2+(5-2)^2} = 3 \end{align*}
\begin{align*} BC \amp = \sqrt{(2-5)^2+(5-1)^2} = 5\\ B'C' \amp = \sqrt{(-5+1)^2+(2-5)^2} = 5 \end{align*}
\begin{align*} CA \amp = \sqrt{(2-2)^2+(1-5)^2} = 4\\ C'A' \amp = \sqrt{(-1+5)^2+(2-2)^2} = 4 \end{align*}
Each pair of corresponding sides is equal: \(AB=A'B'\text{,}\) \(BC=B'C'\) and \(CA=C'A'\text{.}\)
Therefore, by SSS, \(\Delta ABC \equiv \Delta A'B'C'\text{.}\) The rotation has repositioned the triangle without changing its size or shape, so the object and image are directly congruent.

Example 2.3.51.

Triangle \(ABC\) with vertices \(A(2,2)\text{,}\) \(B(5,3)\) and \(C(3,5)\) is rotated \(180^\circ\) about the point \(D(1,1)\text{.}\) Find the image and verify that the two triangles are congruent.
Solution.
A rotation of \(180^\circ\) about \(D(a,b)\) maps \((x,y)\) to \((2a-x,\;2b-y)\text{.}\) With \(a=1\text{,}\) \(b=1\) this is \((x,y)\mapsto(2-x,\;2-y)\text{:}\)
\begin{equation*} A(2,2)\mapsto A'(0,0)\text{,} \end{equation*}
\begin{equation*} \qquad B(5,3)\mapsto B'(-3,-1)\text{,} \end{equation*}
\begin{equation*} \qquad C(3,5)\mapsto C'(-1,-3). \end{equation*}
Checking corresponding sides with the distance formula:
\begin{align*} AB \amp = \sqrt{3^2+1^2} = \sqrt{10}\\ A'B' \amp = \sqrt{(-3)^2+(-1)^2} = \sqrt{10} \end{align*}
\begin{align*} BC \amp = \sqrt{(-2)^2+2^2} = \sqrt{8}\\ B'C' \amp = \sqrt{2^2+(-2)^2} = \sqrt{8} \end{align*}
\begin{align*} CA \amp = \sqrt{(-1)^2+(-3)^2} = \sqrt{10}\\ C'A' \amp = \sqrt{1^2+3^2} = \sqrt{10} \end{align*}
Since \(AB=A'B'\text{,}\) \(BC=B'C'\) and \(CA=C'A'\text{,}\) the triangles are congruent by SSS. Note that congruence holds even though the centre of rotation is \(D(1,1)\text{,}\) not the origin: rotation about any point preserves shape and size.

Example 2.3.52.

Triangle \(ABC\) has vertices \(A(1,2)\text{,}\) \(B(4,2)\) and \(C(1,4)\text{.}\) It is rotated \(-90^\circ\) (that is, \(90^\circ\) clockwise) about the origin. Find the image and deduce whether \(\Delta ABC\) and its image are congruent.
Solution.
Under a \(-90^\circ\) rotation about the origin, \((x,y)\mapsto(y,-x)\text{:}\)
\begin{equation*} A(1,2)\mapsto A'(2,-1)\text{,} \end{equation*}
\begin{equation*} \qquad B(4,2)\mapsto B'(2,-4)\text{,} \end{equation*}
\begin{equation*} \qquad C(1,4)\mapsto C'(4,-1). \end{equation*}
Comparing corresponding sides:
\begin{align*} AB \amp = \sqrt{3^2+0^2} = 3, \amp A'B' \amp = \sqrt{0^2+(-3)^2} = 3;\\ BC \amp = \sqrt{(-3)^2+2^2} = \sqrt{13}, \amp B'C' \amp = \sqrt{2^2+3^2} = \sqrt{13};\\ CA \amp = \sqrt{0^2+2^2} = 2, \amp C'A' \amp = \sqrt{(-2)^2+0^2} = 2. \end{align*}
All three pairs of corresponding sides are equal, so \(\Delta ABC \equiv \Delta A'B'C'\) by SSS. A clockwise rotation is just as rigid as an anticlockwise one, so the image is congruent to the object.

Checkpoint 2.3.53.

Exercises Exercises

1.

Triangle \(ABC\) has vertices \(A(1,1)\text{,}\) \(B(4,2)\) and \(C(2,4)\text{.}\) It is rotated \(90^\circ\) anticlockwise about the origin, giving the image \(A'(-1,1)\text{,}\) \(B'(-2,4)\text{,}\) \(C'(-4,2)\text{.}\)
  1. Use the distance formula to find the lengths of \(AB\text{,}\) \(BC\) and \(CA\text{.}\)
  2. Find the lengths of the corresponding sides \(A'B'\text{,}\) \(B'C'\) and \(C'A'\text{.}\)
  3. Hence explain why \(\Delta ABC\) and \(\Delta A'B'C'\) are congruent, stating the congruence criterion you have used.
Answer.
  1. \begin{align*} AB \amp = \sqrt{(4-1)^2+(2-1)^2} = \sqrt{10} \end{align*}
    \begin{align*} BC \amp = \sqrt{(2-4)^2+(4-2)^2} = \sqrt{8} \end{align*}
    \begin{align*} CA \amp = \sqrt{(1-2)^2+(1-4)^2} = \sqrt{10} \end{align*}
  2. \begin{align*} A'B' \amp = \sqrt{(-2+1)^2+(4-1)^2} = \sqrt{10} \end{align*}
    \begin{align*} B'C' \amp = \sqrt{(-4+2)^2+(2-4)^2} = \sqrt{8} \end{align*}
    \begin{align*} C'A' \amp = \sqrt{(-1+4)^2+(1-2)^2} = \sqrt{10} \end{align*}
  3. Each pair of corresponding sides is equal (\(AB=A'B'\text{,}\) \(BC=B'C'\text{,}\) \(CA=C'A'\)), so \(\Delta ABC \equiv \Delta A'B'C'\) by SSS. Rotation is a rigid transformation, so it preserves lengths and the image is congruent to the object.

2.

Triangle \(PQR\) with vertices \(P(1,2)\text{,}\) \(Q(3,2)\) and \(R(2,5)\) is rotated \(180^\circ\) about the point \(D(-1,0)\text{.}\)
  1. Using the rule \((x,y)\mapsto(2a-x,\;2b-y)\) for a \(180^\circ\) rotation about \((a,b)\text{,}\) find the coordinates of the image \(P'Q'R'\text{.}\)
  2. Show that one pair of corresponding sides has equal length.
  3. Deduce whether \(\Delta PQR\) and \(\Delta P'Q'R'\) are congruent, and give a reason.
Answer.
  1. With \(a=-1\text{,}\) \(b=0\text{,}\) the rule is \((x,y)\mapsto(-2-x,\;-y)\text{.}\) So \(P'(-3,-2)\text{,}\) \(Q'(-5,-2)\text{,}\) \(R'(-4,-5)\text{.}\)
  2. For example, \(PQ=\sqrt{(3-1)^2+(2-2)^2}=2\) and \(P'Q'=\sqrt{(-5+3)^2+(-2+2)^2}=2\text{,}\) so \(PQ=P'Q'\text{.}\) (Similarly \(QR=RP=\sqrt{10}\) and \(Q'R'=R'P'=\sqrt{10}\text{.}\))
  3. All corresponding sides are equal, so \(\Delta PQR \equiv \Delta P'Q'R'\) by SSS. A \(180^\circ\) rotation is a rigid transformation about the point \(D\text{,}\) so the object and image are congruent even though \(D\) is not the origin.

3.

Rotate triangle P(1,1), Q(4,1), R(1,3) 90° anticlockwise about the origin. Find the coordinates of P’, Q’, R’. Verify that ΔPQR ≡ ΔP’Q’R’ by comparing side lengths.
Answer.
  1. P’(−1,1), Q’(−1,4), R’(−3,1)
  2. PQ=3, PR=2, QR=√10; P’Q’=3, P’R’=2, Q’R’=√10, so congruent by SSS

4.

A point A(4,2) is rotated about the origin to A’(−2,4). Determine the angle and direction of rotation. Show that the rotation is an isometry.
Answer.
  1. The rotation is 90° anticlockwise about the origin
  2. OA=√20=OA’, so distances preserved; rotation is an isometry (rigid transformation)
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