Constructions
http://www.mathopenref.com/tocs/constructionstoc.html
Rigid Motion: Geometry Transformations
http://www.mathopenref.com/tocs/transformstoc.html
http://www.mathsisfun.com/geometry/transformations.html
video: http://www.youtube.com/watch?v=cSw2VNJqDTY
Identify and justify congruence via transformations
http://www.mathopenref.com/tocs/congruencetoc.html
Geometry transformations
TRANSFORMATIONS
Students should be able to apply transformations to different objects. They should describe transformations as functions that
take points in the plane as inputs and give other points as outputs – making connections to functions and the transformations
they have done on functions. Students should also make comparisons between transformations that preserve distance and
angle to those that do not. They should be able to predict whether or not a transformation will maintain the congruence
between the pre-image and image – i.e. the input object is congruent to the output object.
NOTE: Using software that allows students to manipulate objects and perform these transformations is highly
recommended. For example: Geometer’s Sketchpad, Geogebra and other programs. These often come with materials,
diagrams that the software company has already created to facilitate explorations of these topics.
TRANSFORMATION TERMINOLOGY
pre-image: original figure before transformation; input
into the transformation function
image: figure after transformation; output from the
transformation function
pre-image
A
B
A
B
C
C
A'
A'
B'
C'
C'
B'
image
TRANSFORMATIONS (rigid motion transformations)
Rigid Motion Transformations: preserve congruence
 reflection – flip
o The reflection of an object is called its image. If the original object (the pre-image) was labeled with letters, such
as polygon ABCD, the image may be labeled with the same letters followed by a prime symbol, A'B'C'D'.
o The line (where a mirror may be placed) is called the line of reflection. The distance from a point to the line of
reflection is the same as the distance from the point's image to the line of reflection.
o A reflection can be thought of as folding and "flipping" an object over the line of reflection.
o Connecting this to functions – the pre-image is the input to the transformational function and the image is the
output.
B
B'
A
Notice that the pre-image B and the image B' define a
line that is perpendicular to the line of reflection the
point of intersection is also the midpoint of BB ' .
A'
D
D'
C
C'
TRANSFORMATIONS (rigid motion transformations - continued)
Rigid Motion Transformations: preserve congruence
 translation – slide
o A translation "slides" an object a fixed distance in a specified direction. The original object and its translation have
the
same shape and size, and they face in the same direction, they are oriented the same way. The word "translate" in
Latin means "carried across".
o Translations move a point a specified distance along a line parallel to a specified line.
o If the pre-image (input) was polygon ABCDE, then the image after a translation has taken place would be polygon
A'B'C'D'E'.
B
C
m
Think of polygon ABCDE as sliding two inches to
the right and one inch down. Its new position is
labeled A'B'C'D'E'.
A
D
B'
E
C'
Notice that the lines m, CC ' and EE ' are parallel.
A'
D'
E'
B'
C'
TRANSFORMATIONS (rigid motion transformations - continued)
Rigid Motion Transformations: preserve congruence
 rotation – A rotation turns an object around a point. Rotations can occur in either a clockwise (to the right) or
counterclockwise (to the left) direction.
o A positive angle of rotation turns the figure counterclockwise, a negative angle of rotation turns the figure
clockwise.
o Rotations move objects along a circular arc with a specified center through a specified angle.
o Notice that a rotation does not change the size of the figure.
C
D
A'
E'
D'
E
B
This rotation of the pre-image polygon
ABCDE is 90° counterclockwise. The
output is the image A'B'C'D'E'.
TRANSFORMATIONS (non-rigid motion transformations --do not preserve congruence)
 dilation – this is not covered in Secondary Mathematics 1

horizontal stretch – a stretch in which a plane figure is distorted horizontally.
o The size and shape of the figure is changed so the image and pre-image are no longer congruent.
Y
Y'
X

X'
Z
Z'
vertical stretch – a stretch in which a plane figure is distorted vertically.
o The size and shape of the figure is changed so the image and pre-image are no longer congruent.
Y'
Y
X
X'
Z
Z'
EXAMPLE (reflection in the coordinate plane)
NOTE: reflection in the origin is the same thing as a 180° rotation about the origin.
Reflections in the Coordinate Plane
Reflection
x-axis
y-axis
origin
y=x
Pre-image
to Image
How to
find
coordinates
(a, b) → (a, -b)
(a, b) → (-a, b)
(a, b) → (-a, -b)
(a, b) → (b, a)
Multiply the y-coordinate
by -1.
Multiply the x-coordinate
by -1.
Multiply both coordinates by
-1.
Interchange the
x- and y-coordinates.
A (1,
Example
B (-3,
1)
B' (-3, 1)
A (2,
3)
A' (-3,
2)
A (3,
2)
A (3,
B' (-3,
B (-
B (3, A' (2,
3)
B' (-1, - B (1, 2)
2)
A'
B' (2, -
A' (-3, y=
EXAMPLE (transformation function)
Here is an example of a transformation function. The original triangle that is graphed is the input or pre-image. The points for the
vertices of the triangle are used as input into the function and the result is a new triangle (or output).
Graph the triangle and its image under the given translation.
∆EFG with vertices E(0, -4), F(-4, -4), and G(0, 2) under the translation (x, y) → (x + 2, y – 1)
G
G'
E(0 + 2, -4 – 1) = E'(2, -5)
F(-4 + 2, -4 – 1) = F'(-2, -5)
G(0 + 2, 2 – 1) = G'(2, 1)
E
F
F'
E'
NOTE: A composition of reflections across two parallel lines (or across any even number of parallel lines) is equivalent to a translation.
A composition of reflections across three parallel lines (or across any odd number of parallel lines) is equivalent to a single reflection. A
composition of reflections over intersecting lines is equivalent to a rotation.
CONGRUENCE
Congruence is defined in terms of rigid motion transformations which preserve distance and angle size. Students should understand that
when translating, rotating or reflecting a figure, the pre-image will be congruent to the image and be able to explain that this means that
the corresponding parts of the figures are congruent and therefore the figures are congruent. Further exploration of this definition of
congruence should lead to an understanding of why ASA, SAS and SSS guarantee the congruence of two triangles.
The emphasis should be on using technology or other models to develop deep understanding. Students should be able to demonstrate
and articulate these ideas.
Theorems to know and understand about congruence:
 CPCTC – Corresponding Parts of Congruent Triangles are Congruent
 ASA – Angle Side Angle: Where the side used is the side in between the two given angles.
ASA

SAS – Side Angle Side: Where the angle used is the angle between the two given sides.
SAS

SSS – Side Side Side
SSS
Geometry Constructions Help
PRECISE DEFINITIONS
Students should know precise (be specific) definitions of the following terms.
angle: the union of two circle: A plane
perpendicular line: two
rays that have the same
figure with all points segments, rays, or lines that
endpoint.
the same distance
intersect to form a right angle
from a fixed point
(90 degree angle.
called a center.
A
P
B
parallel line: two or
more coplanar lines
that have no points in
common or are
identical (e.g., the
same line).
line segment: a line
segment is a part of a
line that is bounded by
two end points, and
contains every point on
the line between its end
points
B
A
90°
θ
Remember a point is not defined because it is a location. A line is made up of points and has no thickness or width and goes on
forever. Distance along a line is not measureable. We can measure the distance between two points on a line.
Explanation of distance around a circular arc: This is rotating an object about a point (specifically the origin) clockwise or
counterclockwise i.e. 90°, 180°, 270°, etc. The path the object traces will create a circular arc.
CONSTRUCTIONS
Copy a segment:
Start: Start with a line segment PQ that we will copy.
Step 1: Mark a point R that will be one endpoint of the new line segment.
Q
P
Step 2: Set the compass point on the point P of the line segment to be copied.
Step 3: Adjust the compass width to the point Q. The compass width is now equal to the
length of the line segment PQ.
Step 4: Without changing the compass width, place the compass point on point R on the
line you drew in step 1.
Step 5: Without changing the compass width, draw an arc roughly where the other endpoint
will be.
S
R
Step 6: Pick a point S on the arc that will be the other endpoint of the new line segment.
Step 7: Draw a line from R to S.
Done: RS is equal in length (congruent to) PQ .
Copy an angle
Start: Start with an angle BAC that we will copy.
B
Step 1: Make a point P that will be the vertex of the new angle.
Step 2: From P, draw a ray PQ. This will become one side of the new angle.
 This ray can go off in any direction.
 It does not have to be parallel to anything else.
 It does not have to be the same length as AC or AB.
Step 3: Place the compass on point A, set to any convenient width.
J
K
C
A
Step 4: Draw an arc across both sides of the angle--create the points J and K as shown.
R
Step 5: Without changing the compass width, place the compass point on P and draw a
similar arc there, creating point M as shown.
Step 6: Set the compass on K and adjust its width to point J.
L
Step 7: Without changing the compass width, move the compass to M and draw an arc
across the first one, creating point L where they cross.
Step 8: Draw a ray PR from P through L and onwards--xact length is not important.
Done: RPQ is congruent (equal in measure) to angle BAC .
M
P
Q
Bisect an angle
Start: Start with angle PQR that we will bisect.
Step 1: Place the compass point on the angle's vertex Q.
P
Step 2: Adjust the compass to a medium wide setting. The exact width is not
important.
Step 3: Without changing the compass width, draw an arc across each leg of the
angle.
Step 4: The compass width can be changed here if desired. Recommended: Leave
it the same.
Step 5: Place the compass on the point where one arc crosses a leg and draw an arc
in the interior of the angle.
Q
Step 6: Without changing the compass setting repeat for the other leg so that the
two arcs cross.
R
Step 7: Using a straightedge, draw a line from the vertex to the point where the arcs
cross.
Done: This is the bisector of the angle ÐPQR .
Perpendicular Bisectors:
Start: Start with a line segment PQ.
Step 1: Place the compass on one end of the line segment.
Step 2: Set the compass width to approximately two thirds the line length.
The actual width does not matter.
Step 3: Without changing the compass width, draw an arc above and
below the line.
J
P
Q
Step 4: Again without changing the compass width, place the compass
point on the other end of the line. Draw an arc above and below
the line so that the arcs cross the first two.
Step 5: Using a straightedge, draw a line between the points where the
arcs intersect.
Done: This line is perpendicular to the first line and bisects it (cuts it at
the exact midpoint of the line). Therefore, PJ  JQ
Parallel Lines:
Start: Start with a line segment PQ and a point R off the line.
Step 1: Draw a transverse line through R and across the line PQ at an
angle, forming the point J where it intersects the line PQ. The
exact angle is not important.
R
S
Step 2: With the compass width set to about half the distance between R
and J, place the point on J, and draw an arc across both lines.
Step 3: Without adjusting the compass width, move the compass to R and
draw a similar arc to the one in step 2.
Step 4: Set compass width to the distance between where the lower arc
crosses the two lines. Move the compass to where the upper arc
crosses the transverse line and draw an arc across the upper arc,
forming point S.
Step 5: Draw a straight line through points R and S.
suur
sur
Done: RS is parallel to PQ .
P
J
Q
Construct a perpendicular from a line ghrough a point
Start: Start with a line and point R which is not on that line.
R
Step 1: Place the compass on the given external point R.
Step 2: Set the compass width to approximately 50% more than the distance
to the line. The actual width does not matter.
Step 3: Draw an arc across the line on each side of R, making sure not to
adjust the compass width in between.
J
P
Step 4: At this point, you can adjust the compass width. Recommended:
leave it as is. From each point P and Q, draw an arc below the line
so that the arcs cross.
Q
Step 5: Align a straightedge between R and the point where the arcs intersect.
Draw the perpendicular line from R to the line, or beyond if you
wish.
Done: This line is perpendicular to the first line and passes through the point
R.
Construct a perpendicular from a line at a point
Start: Start with a line and point K on that line.
R
Step 1: Set the compass width to a medium setting. The actual width does
not matter.
Step 2: Without changing the compass width, mark a short arc on the line at
each side of the point K, forming the points P and Q. These two
points are thus the same distance from K.
Step 3: Increase the compass to almost double its width (again the exact
setting is not important).
Step 4: From P, mark off a short arc above K.
90°
Step 5: Without changing the compass width repeat from the point Q so that
the two arcs cross each other, creating the point R.
Q
K
P
Step 6: Using the straight edge, draw a line from K to where the arcs cross.
Done: KR just drawn is a perpendicular to PQ at K.
Construct an equilateral triangle in a circle
Start: Start with the line segment AB which is the length of the sides of the
desired equilateral triangle.
A
R
Step 1: Pick a point P that will be one vertex of the finished triangle.
Step 2: Place the point of the compass on the point A and set it's drawing
end to point B. The compass is now set to the length of the sides of
the finished triangle. Do not change it from now on.
Step 3: Place the point of the compass on the point A and set it's drawing
end to point B. The compass is now set to the length of the sides of
the finished triangle. Do not change it from now on.
Step 4: On one of the arcs, mark a point Q that will be a second vertex of
the triangle. It does not matter which arc you pick, or where on the
arc you draw the point.
Step 5: Place the compass point on Q and draw an arc that crosses the other
arc, creating point R.
Done: The ∆PQR is an equilateral triangle. Its side length is equal to the
distance AB.
B
P
Q
Construct an equilateral triangle in a circle
Start: Start with the line segment AB which is the length of the sides of the
desired equilateral triangle.
A
R
Step 1: Pick a point P that will be one vertex of the finished triangle.
Step 2: Place the point of the compass on the point A and set it's drawing end to
point B. The compass is now set to the length of the sides of the
finished triangle. Do not change it from now on.
Step 3: Place the point of the compass on the point A and set it's drawing end to
point B. The compass is now set to the length of the sides of the
finished triangle. Do not change it from now on.
B
Step 4: On one of the arcs, mark a point Q that will be a second vertex of the
triangle. It does not matter which arc you pick, or where on the arc you
draw the point.
Q
P
Step 5: Place the compass point on Q and draw an arc that crosses the other arc,
creating point R.
Done: The ∆PQR is an equilateral triangle. Its side length is equal to the
distance AB.
Construct a hexagon in a circle
Start: Start with a given circle, center O.
Step 1: Mark a point anywhere on the circle. This will be the first vertex of the hexagon.
Step 2: Set the compass on this point and set the width of the compass to the center of the
circle. The compass is now set to the radius of the circle.
Step 3: Make an arc across the circle. This will be the next vertex of the hexagon. (It turns
out that the side length of a hexagon is equal to its circumradius - the distance
from the center to a vertex).
O
Step 4: Move the compass on to the next vertex and draw another arc. This is the third
vertex of the hexagon.
Step 5: Continue in this way until you have all six vertices.
Step 6: Draw a line between each successive pairs of vertices, for a total of six lines.
Done: These lines form a regular hexagon inscribed in the given circle.
Construct a square
Start: Start with line segment PQ of length s.
Step 1: Extend the line segment past Q.
S
R
P
Q
Step 2: Erect the perpendicular to ray PQ at Q.
Step 3: Using the line drawn in the previous step, mark off a line
segment of length s such that one of its endpoints is Q . Label
the other endpoint as R.
Step 4: Draw an arc of the circle with center P and radius PQ.
Step 5: Draw an arc of the circle with center R and radius QR to find the
point S where it intersects the arc from the previous step.
Step 6: Draw the square PQRS.
Done: PS=PQ=QR=QS, yielding that PQRS is a rhombus. Since PQR
is a right angle, it follows that PQRS is a square.