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Symmetries of a quadrilateral II

Alignments to Content Standards: G-CO.A.3


There is exactly one reflection and no rotation that sends the convex quadrilateral ABCD onto itself. What shape(s) could quadrilateral ABCD be? Explain.

IM Commentary

This task examines quadrilaterals from the point of view of rigid motions and complements www.illustrativemathematics.org/illustrations/1471. A complete and justified answer to this problem is a challenging task which requires dividing into and analyzing several possible cases for the line $\ell$ of reflection:

  • $\ell$ contains no vertices of $ABCD$,
  • $\ell$ contains one vertex of $ABCD$,
  • $\ell$ contains two vertices of $ABCD$.

In the first case, $ABCD$ is a regular trapezoid (also called an isosceles trapezoid). The second case is not possible and in the third case $ABCD$ is a kite. As is the case for triangles, it is interesting that most of the different types of quadrilaterals can be characterized in terms of their symmetries.

If appropriate geometry software is available, this task provides an ideal opportunity for students to engage in MP5, Use Appropriate Tools Strategically, as students can experiment with different quadrilaterals in order to determine which ones have reflective symmetry. Once they have reached a conclusion about what type of quadrilateral $ABCD$ is, they will need to Reason Abstractly and Quantitatively (MP2) in order to show that $ABCD$ is either an isosceles trapezoid or kite. Reasoning to show that the line of reflection cannot pass through just one vertex of $ABCD$ also requires careful reasoning. The task is ideally suited for group work both because of it complexity and because of its natural division into cases.


Suppose $\ell$ is a line of symmetry of $ABCD$ and let $r_\ell$ be the transformation that reflects over $\ell$. If $\ell$ contains one vertex of $ABCD$ then it must contain a second vertex. To see why, note that $r_\ell$ maps vertices of $ABCD$ to vertices of $ABCD$. Suppose for simplicity that $A$ is on $\ell$ but $B,C,D$ are not. We have $r_\ell(A) = A$. Since $B$ does not lie on $\ell$ this means that $r_\ell(B) \neq B$. So we must have $r_\ell(B) = C$ or $r_\ell(B) = D$. In the former case then $r_\ell(D) = D$ which is not possible since then $r_\ell$ would send two vertices to themselves. Similarly, if $r_\ell(B) = D$ then $r_\ell(C) = C$ which is also not possible.

This leaves two cases to treat: the situation where $\ell$ contains no vertices of $ABCD$ and the situation where it contains two of these vertices. We first deal with the case where $\ell$ contains no vertices of $ABCD$. Since certainly a line of symmetry cannot be completely outside the quadrilateral, we know in this case that two vertices will lie on one side of $\ell$ and the other two vertices will lie on the other. An example is pictured below:


We have that $\ell$ is the perpendicular bisector of $\overline{AB}$ and also the perpendicular bisector of $\overline{CD}$. This means that $\overleftrightarrow{AB}$ is parallel to $\overleftrightarrow{CD}$ since they are both perpendicular to $\ell$. Thus $ABCD$ is a trapezoid. Note that $r_\ell(A) = B$, $r_\ell(B) = A$, $r_\ell(C) = D$, and $r_\ell(D) = C$ so $r_\ell$ interchanges $\overline{DA}$ and $\overline{CB}$. Thus $\overline{DA}$ and $\overline{CB}$ are congruent and $ABCD$ is a regular trapezoid. It cannot be a rectangle or else there is a second reflection that maps $ABCD$ to itself. So we have in addition $|AB| \neq |CD|$.

Below is a picture of the second case where $\ell$ passes through two vertices of $ABCD$:


Since $\ell$ contains two vertices of $ABCD$ the other two must be reflections of one another about $\ell$. In the picture $\ell$ goes through $A$ and $C$ and $D = r_\ell(B)$. So $r_\ell$ interchanges $\overline{AD}$ and $\overline{AB}$ and also interchanges $\overline{CB}$ and $\overline{CD}$. So if $|AB| = |BC|$ then $ABCD$ is a rhombus and has a second line of reflective symmetry, namely $\overleftrightarrow{BD}$. Quadrilateral $ABCD$ is what is called a kite. Note that we have implicitly used the fact that $ABCD$ is convex: without this hypothesis, the vertex $C$ could be reflected over $\overleftrightarrow{BD}$, giving a non convex quadrilateral with exactly one line of symmetry.