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Reflections and Isosceles Triangles


Alignments to Content Standards: G-CO.B

Task

Suppose ABC is an isosceles triangle with side \overline{AB} congruent to side \overline{AC} as pictured below:

Reflection1_fceae0123a8f22eef307ebb42f8ab930

Also pictured above is the midpoint D of side \overline{BC}. Let L denote line \overleftrightarrow{AD} and let r_L denote the map of the plane determined by reflection about the line L.

  1. Show that r_L(A) = A, r_L(B) = C, and r_L(C) = B.
  2. Using part (a) show that r_L maps the triangle ABC to itself.
  3. Suppose EFG is a triangle in the plane and that there is a line M so that reflection about M maps triangle EFG to itself. Show that triangle EFG is an isosceles triangle.
  4. Using parts (b) and (c), show that a triangle is isosceles if and only if there is a reflection of the plane that preserves the triangle.

IM Commentary

This activity is one in a series of tasks using rigid transformations of the plane to explore symmetries of classes of triangles, with this task in particular focussing on the class of isosceles triangles. Students will have previously seen similar material in earlier grades (e.g., standard 4.G.3: "Recognize a line of symmetry for a two-dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify line-symmetric figures and draw lines of symmetry."), and this task is a similar exploration with a level of mathematical precision more appropriate for high schools students. In particular, the task has students link their intuitive notions of symmetries of a triangle with statements proving that said triangle is unmoved by applying certain rigid transformations.

The mechanism through which students do this is via congruence proofs, so the task illustrates the general relationship between congruence and rigid transformations, as fits naturally under the cluster "understand congruence in terms of rigid motions." At the high school level, students will need to use the definition of reflections and the teacher may wish to recall this before having students work on the problem. The instructor may wish to discuss with students how to draw an appropriate picture for part (b) and part (c) as this will be essential in order to make progress on these problems. In addition, it may be necessary to recall the definition of reflection about a line L: reflection about L maps each point on L to itself and if P is not on L then reflection about L maps P to Q when L is the perpendicular bisector of segment PQ as pictured below:

Reflection2_22b53bcc7969a36b7e46571b0f06ed12

This task is mainly intended for instructional purposes: for students with a strong visual sense of geometry, rigid motions of the plane provide a rich source of ideas which can be used to grasp many important geometric ideas.

Solution

  1. Since the reflection over line L leaves each point on L fixed, we have r_L(A) = A since A lies on L. Since B and C do not lie on L, in order to show that r_L(B) = C and r_L(C) = B we need to show that L is the perpendicular bisector of segment \overline{BC}.

    Note that we are given that segment \overline{AB} is congruent to segment \overline{AC}. Since D is the midpoint of segment \overline{BC} this means that segment \overline{BD} is congruent to segment \overline{CD}. Finally segment \overline{DA} is congruent to segment \overline{DA}. So this means, by SSS, that triangle ABD is congruent to triangle ACD. Since angles BDA and CDA together add up to 180 degrees and they are congruent it follows that they are both right angles. Therefore L is perpendicular to segment \overline{AB} and it is the bisector of \overline{AB} by assumption.

  2. From part (a) we know that r_L(A) = A, r_L(B) = C, and r_L(C) = B. Reflections map line segments to line segments so it follows that r_L maps side \overline{AB} to side \overline{AC}, side \overline{AC} to side \overline{AB} and side BC to itself. Hence r_L maps triangle ABC to itself. Note that although r_L maps side \overline{BC} of triangle ABC to itself, the map interchanges vertices B and C so the orientation of this segment is reversed.

  3. We are given that r_M maps triangle EFG to itself. Suppose we look at r_M(E). Since r_M preserves triangle EFG it must interchange the vertices of E, F, G in some way. One possibility would be that it leaves them all alone, that is E maps to E, F maps to F, and G maps to G. This can not happen, however, because a reflection of the plane can not leave \triangle EFG in its original position. If two of the vertices are preserved by r_M then the third must be also as there is no freedom to send it to either of the two fixed vertices. So this means that the reflection r_M must leave exactly one vertex of \triangle EFG fixed and it must exchange the other two vertices. For simplicity, let's assume that r_M(E) = E, r_M(F) = G, and r_M(G) = F. This situation is pictured below:

    Trireflection_e3740e81dcf00059a5605caa36efd5d8

    Since r_M preserves distances this means that |EF| = |r_M(EG)| = |EG| and so \triangle EFG is isosceles.

  4. Part (b) shows that for each isosceles triangle ABC there is at least one line L so that r_L maps triangle ABC to itself. Part (c) shows that if there is a line L so that r_L maps triangle ABC to itself then triangle ABC is isosceles. Putting these two statements together we see that triangle ABC is isosceles if and only if there is a line L so that r_L maps triangle ABC to itself.