Introduction to
The Geometer’s Sketchpad
Open the program, click in the white space to close the program icon. You should have a new
sketch showing.
Down the left are the tools and across the top is a drop down menu of options.
look at both of these a bit more closely.
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Exercise 1
Now let’s check out that connecting the ends of a diameter to a point on the circle creates a right
triangle.
Here is a list of instructions:
To Check on that Right Triangle thing:
Click on the circle tool, click and drag on the sketch. You have just created a circle with the
circle tool. Click in the white space to unselect the circle and points.
Create a diameter by using the pointer tool to click on the circle point and the center (in that
order) and then going to the drop down menu “Construct” and selecting “Ray” while just the 2
points are highlighted . Click in the white space
Click where the ray hits the circle to put a point at the other end of the diameter. Click in the
white space.
Now select the ray and go to “Display” and “Hide Ray”.
Now with the pointer tool, put a point on the circle. Then deselect everything.
Then select the 3 points: right end, left end and circle point. Go to “Construct” and choose
“Segments”. You now have your triangle circumscribed by the circle.
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You should have something that looks like this.
Now click the a leg point, the vertex, and the other leg point in that order and go to “Measure”
and then measure the angle.
See how “Angle” is bold?
Click it.
Notice that GSP will name your points for you! You may move the angle measure anywhere on
the sketch that you want by using the point tool.
Now let’s animate the 90 degree angle. Clear all selections, click on the 90 degree vertex with
your point. Go to display and select “Animate Point”. It’s going to circle around and you’ll note
that the measure of the angle is steady…it’s 90 degrees no matter which circle point the vertex is
using.
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Your screen should look something like this:
Note that this is an illustration and NOT a proof!
Put notes about the construction here:
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Exercise 2
You will have an axiom that states if a pair of angles are a linear pair, then they are supplements.
Now a linear pair of angles share an adjacent leg. They look like this:
Can you get something like this on your sketch? Close the old sketch and don’t save it (small
blue box upper right). Get a new sketch under “File”. Use the segment tool to start.
Now, an important point. You do NOT want to make right angles with your linear pair. Yes,
right angles are ONE of an infinite number of angle measures…but they are a special case for
that very reason! Always choose the most typical or general case when you are illustrating
something.
Construction notes:
Now let’s get a sum and then move the ray. Deselect everything. Go to “Number” and select
“Calculator”. Click on the first angle measure, the plus sign, and the second angle measure.
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Then click on “ok”. The sum will, of course, be 180 degrees.
Complements
Supplements
Mnemonic:
90
180
Alpha order, numerical order smallest to largest
Now move the ray. Deselect everything and select the end point of the shared leg. Toggle it.
Note that the angle measures change when you do that BUT the sum is steady. Again, an
illustration but NOT a proof.
Construction notes:
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Exercise 3
Let’s discuss the incircle. This is a special circle, inside an arbitrary triangle, tangent to each
side of the triangle. As you can imagine, this has a very special construction.
First put a triangle on your new sketch. Build it with segments or by putting on 3 points and
connecting them with segments. Then construct angle bisectors to each vertex. Put a point at the
intersection of the angle bisectors and hide the segments.
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Now you really could have gotten away with only doing 2 bisectors BUT I wanted to show you
that all three intersect at one point and to give you more practice on getting the bisectors.
Next, select the incenter (the point of intersection) and one side of the triangle and construct a
perpendicular line. Put a point at the intersection of the line and the triangle’s side. Hide the
perpendicular line.
Then select the incenter and the new point, in that order, and construct a circle from center and
point. You should have a circle in the triangle. This is the biggest circle you can put into the
triangle. It is tangent to each side. Move the vertices of the triangle around and notice that the
circle stays inside!
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Construction notes:
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Exercise 4
A 3 – 4 – 5 right triangle and one similar to it.
Put a point on a new sketch. With the point still selected go to the “Transform” tab, click on
“Translate” and put in 3 cm and 90 degrees. Then click on “Translate”.
Clear the sketch. Select your original point and translate it 4 cm at 0 degrees. Then connect the
vertices to create a right triangle.
Let’s name the points. Select the points (in order)…the original point, the first translated point,
then the second translated point.
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Noticed that the translated points are named for the original point. Let’s rename the apex, or
highest point, “B”. Use the alphabet tool, put the little hand on the label “A’” and double click.
Change the name and then rename the rightmost point on the base “C”.
Then clear. Select, in order, A, B, and C and go to the Transform menu, select dilate. Make sure
“about a fixed point” is marked and that the fraction is ½. Click dilate. Connect the new points
with a segment and name them D on the vertical and E on the horizontal.
Measure the legs to check that they are indeed one-half the length of the original legs.
Now let’s move that smaller triangle to the right. Select A, D, and E. Go to the Transformations
tab and select “Translate”…move it over 6 cm at 0 degrees. Construct the new smaller triangle
and hide D, E, and the segment. Name the points A, D, and E.
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Construction Notes:
Exercise 5
Creating an equilateral triangle is exceptionally easy to do.
Put a medium sized segment in a new sketch. Select the left endpoint and double click…you
will see a burst that looks like a target. Then select the right endpoint. Without clicking
anything else go to the Transform tab and select rotate. Rotate at 60 degrees…if the change is
not up, put in − 60.
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Then hit rotate and connect the points with segments. Measure the angles to make sure each one
is 60 degrees!
Construction notes:
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Exercise 6
Areas and Altitudes of similar triangles exercise:
A.
3 – 4 – 5 triangles:
In the following 3 – 4 – 5 right triangle, the height is 3 units. In a similar triangle on the right,
the height is 1.5; the scale factor, or constant of proportionality, from left to right is ½.
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3
2.5
1.5
4
2
What is the area of each triangle?
What is the ratio of the area of the smaller triangle on the right to the area of the triangle on the
left?
What is the constant of proportionality or scale factor?
How are these numbers related?
What is the altitude of each triangle? What is the ratio of the smaller to the larger? How is this
related to the constant of proportionality?
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B.
Isosceles right triangles
All isosceles right triangles are similar. Why?
Here are two similar (of course!) isosceles right triangles:
B'
A'
B
A
B''
A''
A'A = 2.00 cm
B'B = 4.00 cm
What is the constant of proportionality from left to right?
What is the area of each triangle?
What is the ratio of the area of the larger triangle on the right to the area of the triangle on the
left?
How are these numbers related?
What is the altitude of each triangle? What is the ratio of the smaller to the larger? How is this
related to the constant of proportionality?
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C.
Here is a pair of arbitrary similar triangles
First:
Make a conjecture about the relationship of the altitudes and the areas?
What do you think is going to be the relationship?
DD' = 6.00 cm
E'
AA' = 4.00 cm
E'D = 3.00 cm
AB' = 2.00 cm
B'C = 1.73 cm
B'
C
A
E'F = 2.60 cm
A'
F
D
D'
B' C is the altitude of AA’B’ and E ' F is the altitude of DD’E’.
Why do you suppose we use the word “height” in the area formula when we really mean
“length”?
What is the scale factor k from AA’B’ to DD’E’?
What are the areas? What is the ratio of the area of AA’B’ to DD’E’?
What about the ratio of the altitudes?
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Summary of area and altitude of similar triangles exercise:
How do you know the triangles are similar?
What is the constant of proportionality?
B''
B
D''
D
C''
A
C
A'
BA = 3.50 cm
AC = 2.50 cm
m BAC = 135.00 
B''A' = 5.25 cm
m B''A'C'' = 135.00 
m B''C''A' = 32.46 
m ACB = 32.46 
Calculate the exact area of the smaller triangle using the trigonometric formula.
What is the area of the triangle on the right – without calculating!
What is the length of A’D” if AD = p cm?
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Exercise 7
The alternate interior angles of a transversal crossing parallel lines are congruent.
Note: A transversal is a line that crosses two other lines whether they are parallel or not…so the
base of a triangle is a transversal crossing the two other legs.
Let’s work with parallel lines. Put a segment across most of the new sketch and put a point not
on the segment. Have them both selected. Go to Construct and click on Parallel Line. Deselect
everything and put a transversal segment across the parallel lines. You need to put points at the
intersection of the transversal and the lines and you need to put points on the parallel line you
created so that you have 3 points to click on for each angle. Make SURE the vertex point is the
second one you click and measure those angles to check.
How did you do?
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Exercise 8
We will continue working with similar triangles by working on spotting similar triangles in some
constructions.
Example: Constructing a congruent angle and similar triangles with any angle
Prove that ADG ~ FEG and that A  F ,
C
D
Note: Figure CAB is an angle. Put an angle
on a new sketch and name it CAB.
E
A
B
G
F
So the construction is:
Take a perpendicular line to one leg of an angle and take a perpendicular to the second leg –
extend them until they intersect. This construction creates similar triangles.
Proof:
DGA  EGF because they’re vertical angles.
ADG  GEF by construction. So by the AA Similarity Theorem ΔADG ~ ΔFEG.
And, in particular note that angle A is congruent to angle F.
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Note the alternate illustration – it doesn’t change the substance of the proof:
Can you figure out how to color in those triangle interiors?
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Similarity Proof Exercise:
Constructing similar triangles with parallel lines.
Prove that ABC ~ EBD.
You are given that AC is parallel to DE .
D
A
B
C
E
The construction is to take parallel lines and cross them with two transversals that intersect in
between the parallel lines.
If AC =
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DE , what do you know about the area of ABC relative to the area of EBD?
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Both of these constructions give you a quick way to construct an angle that is congruent to a
given angle or to construct similar triangles
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Additonal Exercise: Constructing similar isosceles triangles
An easy way to construct a pair of nested similar isosceles triangles follows.
Take an isosceles triangle in which the base is not equal to the sides and the apex angle is less
than 60. (Why does it need to be less than 60°?) Using the base as a side, construct a smaller
interior triangle inside the original one by rotating a ray up from the base until the new angle
created inside one of the base angles is congruent to the apex angle of the original triangle.
A
mBAC = 35.00
Prove that ΔABC ~ ΔBDC
The two triangles share  C.
By construction, A  DBC
So, by the AA Similarity Theorem
D
B
ABC ~BDC.
C
mDBC = 35.00
If AB is 6 cm and BC is 2 cm, how long is DC ?
What is the scale factor from ABC to BCD?
k = 1/3.
So, DC 
1
3
BC 
2
3
How do the areas compare?
Area ΔBCD = 1/9 area ΔABC = 1/9 (1/2) 6(6) sin35°, exactly.
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Exercise 9
Exercise: Constructing similar right triangles with an interior altitude.
Three similar right triangles are created by the altitude from the right angle to the hypotenuse of
any right triangle.
D
How will we construct this?
B
A
mADB = 90.00
C
D
B
A
mADB = 90.00
ΔACD ~ ΔABD
C
They share angle A and both have one right angle. AA Similarity.
D
ΔCBD ~ ΔDBA
They share angle B and both have one
right angle. AA Similarity
B
A
mADB = 90.00
C
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ΔACD ~ ΔDCB
Which is to say, the two smaller triangles
are also similar.
D
Note that  A and  B are complements.
 A = 90° −  B.
B
A
In ΔACD, we have a right angle,  A and
 ADC = 90° −  A =  B.
C
mADB = 90.00
In ΔDCB, we have a right angle and  B.
So, by the AA Similarity Theorem, these are similar.
Thus, by constructing an interior altitude, you get 3 pairs of similar right triangles.
Also note that the following ratios hold:
D
B
A
C
mADB = 90.00
AB BD

BD CB
hypotenuse
shortest leg
AB AD

AD AC
hypotenuse
medium leg
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Exercise 10
The Morley triangle is equilateral.
Create an arbitrary triangle. Measure the sides and angles. Trisect each angle. Chose the
intersection of the trisectors CLOSEST to a side as the vertex of a triangle. Hide the trisectors.
Measure the sides of the Morley triangle. Move the triangle to see what happens to it as you
create special triangles.
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