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# Congruent Segments

Alignments to Content Standards: 8.G.A.2

Line segments AB and CD have the same length. Describe a sequence of reflections that exhibits a congruence between them.

## IM Commentary

Students' first experience with transformations is likely to be with specific shapes like triangles, quadrilaterals, circles, and figures with symmetry. Exhibiting a sequence of transformations that shows that two generic line segments of the same length are congruent is a good way for students to begin thinking about transformations in greater generality.

## Solution

Recall the definition of a reflection: Given a fixed line $L$, a point P is sent to P' so that

• The line through P and P' is perpendicular to $L$, and

• The distance from P to $L$ is equal to the distance from to P' to $L$.

So a reflection that sends P to P' must have a line of reflection that is the perpendicular bisector of the segment PP'.

Create the midpoint M of segment AC and draw the line through M and perpendicular to AC.

Reflect AB across this line. The reflection sends A to C and B to a point we will call B'

If B' happens to land on D, then we are done. If not, create the midpoint of the segment B'D and the line through this midpoint and point C. This line is the perpendicular bisector of B'D (in High School, this should be proved, but for Grade 8 it can be observed. It follows from SSS congruence). Reflect CB' across this line. Since C is on the line it is fixed, and B' is sent to D.

This shows that for any two segments of the same length, one can be mapped to the other by a sequence of at most two reflections.

#### Jason says:

over 1 year

If students get stuck, perhaps you could give a hint along the lines of, "Begin by finding a reflection that sends point A to point C."

#### anmueller says:

over 6 years

Comment on Extension #1:

That is not true. Let X be the intersection of lines AB and CD. If triangle XBD is isosceles with base BD, then a single reflection across the perpendicular bisector of BD will map them also.

Perhaps in simpler terms, if ABCD is an isosceles trapezoid, a single reflection is possible.

Comment on Extension #2:

Consider the triangle BB'D. The arc of rotation is the circumcircle of that triangle. Hence the perpendicular bisector of segment BD also goes through the center of rotation. Therefore we can find the center of rotation as the intersection of the segments AC and BD, i.e. we don't need B' to find it.

Also: If ABCD is a rectangle, a single rotation is still possible. The center of rotation is the intersection of the diagonals. Then a 180 degree rotation will map the segments.

over 6 years

removed