GSP LAB: WORKING WITH THE POINTS OF CONCURRENCY
PART ONE: A Conclusion about the Perpendicular Bisector
1. Create a segment AC.
2. Create the perpendicular bisector of segment AC. (To do this, first construct the
midpoint of segment AC. Then, select the midpoint and select segment AC and use the
Construct menu to create the perpendicular line.)
3. Plot a point anywhere on the perpendicular bisector. Label the midpoint of AC as point
D. Label the point on the perpendicular bisector as E. Your drawing should look
approximately as follows:
4. Select points E and A and use the Measure menu to find the distance between the
points. Repeat for points E and C. Drag point E anywhere along the perpendicular
bisector. You should be noticing that the distance from A to E is always equivalent to
the distance from C to E.
That leads to the following conjecture:
“Any point that lies on the perpendicular bisector of a segment is
equidistant from the endpoints of the segment.”
Now, onward to prove this conjecture!!!
5. To aid in the proof, we can create segments between E and A and between E and C in
our diagram. Now, using the diagram below, let’s identify the given statements (in
other words, let’s list the important constructions that we created in the sketch):
Given: Point D is the midpoint of segment AC
Line ED is perpendicular to segment AC
Now, if we think about what we observed, it seems that point E is always equidistance
from points A and C. So, that is what we will try to prove as stated here:
Prove: Point E is equidistant from points A and C.
Now, carry out the proof!
E
A
D
C
PART TWO: A CONSTRUCTION INVOLING THREE PERPENDICULAR BISECTORS
If all went well in Part One, you have successfully proved that any point on the perpendicular
bisector of a segment is equidistant from the endpoints of the segment.
Now, let’s explore an interesting result in geometry that makes use of three perpendicular bisectors.
1. Using the segment tool, create an acute triangle ABC as shown.
B
C
A
2.
Plot the midpoint of side AC by highlighting side AC and choosing Midpoint from the Construct
menu. Then, highlight side AC and the midpoint and choose Perpendicular Line from the
Construct menu. You have just plotted the perpendicular bisector of side AC. Now, construct
the two other perpendicular bisectors of this triangle.
3. Carefully plot a point at the intersection of the perpendicular bisectors. This point of
intersection of the three perpendicular bisectors of a triangle is called the circumcenter of the
triangle.
4. Using the circle tool, and using the circumcenter as the center, plot a circle that passes through
each of the vertices of the triangle. Make sure to “lock” the control point of the circle onto one
of the vertices of the triangle. Label the center of the circle as point D. Your image should look
approximately as follows:
B
D
A
5.
C
Now, drag any vertex of the triangle to confirm that the circumcenter ALWAYS marks the center
of a circle that passes through all three vertices of the triangle.
6. That was neat! We took a triangle, constructed the three perpendicular bisectors, and noted
that they all met at a point. Then, that point served as the center of a circle that JUST SO
HAPPENED to pass through each of the three vertices of the triangle.
Magic, right?
NO!!! We can explain WHY this happens based upon mathematics and reasoning!
Our goal now is to prove that when we construct a perpendicular bisector for each of the three
sides of a triangle, the point of intersection of those perpendicular bisectors (the circumcenter)
will indeed ALWAYS serve as the center of a circle that passes through the three vertices of the
triangle.
Onward again!!!
7. The diagram of your construction is included again here. I have added cursive lower case letters
to name the lines:
A circle is the set of all points
equidistant from a point called the center.
Take a moment to recall the definition of circle:
So, if we are going to prove that the circle centered at D passes through points A, B and C, then
we are going to need to show that points A, B and C are equidistant from point D. In other
words, we will need to show that AD  BD  CD .
Using the theorem we proved earlier in this lab (that a point on a perpendicular bisector of a
segment is equidistant from the endpoints of the segment), note how we can answer the
following question:
What can we say about point D in relation to points A and B?
Answer: Line l is the perpendicular bisector of segment AB (given). So, any point on line L,
like point D, is equidistant from the endpoints of the segment (namely points A and B).
Recall, we have proven this earlier in the lab! So, we know that the length of DA is the same
as the length of DB. So, we can note that
AD  BD .
Hence, if we draw a circle centered at D, we could make it pass through at least points A and
B now since A and B are the same distance from D. We are already well on our way to
proving that AD  BD  CD . We have already shown that AD  BD !!!
Now, you take over by following through the next set of questions.
a) Similarly, what can we say about point D in relation to points A and C? Justify your
conclusion as modeled above. Note the congruency that results, also as modeled above.
b) What property now allows us to say that AD  BD  CD ?
c)
Note that we have just proven the result! Which choice below best describes what we have
just proven? (Circle one)

Given a circle, a triangle can be formed by connecting any three points on the
circle.

The three perpendicular bisectors of a triangle intersect to form a circle congruent
to the circle that is formed by the three medians of a triangle.

The three perpendicular bisectors of a given triangle intersect at a point that
serves as the center of a circle that passes through each vertex of the triangle.

The three perpendicular bisectors of a circle always intersect to form a triangle.
Note that by explaining how this construction indeed works, we have taken the mystery and magic
out of this process. A good mathematician understands that there is always an explanation based on
logic behind any perceived magic! This, in turn, makes the magicians sad.
PART THREE: A CONJECTURE ABOUT AN ANGLE BISECTOR
1. Use the segment tool to create an angle ABC in a new sketch.
2. Select the three points in order from A to B to C. Then, use the Construct menu to create an
angle bisector.
3. Plot a point D on the angle bisector. Keep point D somewhat close to point B. Your diagram
should look approximately as shown.
A
D
C
B
4. We now wish to find the distance from point D to segment AB. We always measure
distance from a point to a segment along a perpendicular. To do this, follow these steps:
A) Select point D and segment AB and then create a perpendicular line from the Construct
menu.
B) Use the point tool to plot a point where the perpendicular line intersects segment AB.
Label this point as point E.
C) Select the perpendicular line and then hide it (from the Display menu). Make sure not
to hide anything else!
D) Use the segment tool to create a segment from D to E. You should have an image that
looks approximately like the one below:
A
E
D
C
B
E)
Now highlight segment DE and use the Measure menu to find the length. This gives us
the distance from point D (which lies on the angle bisector) to one of the sides of the
angle. (Note again that distance from a point to a line always references the shortest
possible distance from the point to the segment, which is always along the
perpendicular!)
5. We now wish to repeat the process to find the distance between point D and segment BC.
Please do this now. Label the intersection of the perpendicular line and segment BC as
point F. Be sure to measure segment DF as well. When finished, check your image with the
one below:
A
E
D
C
B
F
You should have noticed that the length of segment DE is equal to the length of segment DF.
As a result, we have observed that point D is equidistant from segments AB and BC. (The
assumption is that we are measuring distance along the perpendicular from D to each side
of the angle.) Hence, we make the following conjecture based on what we have observed:
“A point on an angle bisector is equidistant to each side of the angle.”
Onward to prove this!
6.
For the first given, recall that we constructed the angle bisector. So our first given statements
should define the angle bisector. Second, we also created perpendiculars from point D to the
sides of the angle. So, our second set of given statements should define the perpendicular
segments.
Then, define identify what we are trying to prove (see #5 above). Use good symbolism.
Given:
Prove:
A
E
D
C
B
F
Carry out your proof below:
PARTS 4 AND 5 ARE OPTIONAL, BUT WORTHY OF BONUS POINTS!
PART FOUR: THE TRIANGLE INCENTER
1.
Using the segment tool, construct triangle ABC.
B
C
A
2. Construct the three angle bisectors.
3. Carefully plot a point where the angle bisectors intersect. Label the point as D. The intersection
of the three angle bisectors is called the incenter of the triangle. Your sketch should look
approximately like the one below:
B
D
A
C
4.
In the image below, the segments representing the shortest path to each side are shown. Using
the skills you learned in Part Three above, create each of these segments. When finished, your
diagram should look as follows:
B
E
F
D
A
G
C
A) Given that point D is on the angle bisector for angle BAC, which segments can we conclude
are congruent as a result?
B) Given that point D is on the angle bisector for angle ABC, which segments can we conclude
are congruent as a result?
C) Is it right to conclude that Point D is the same distance from all three sides of the triangle?
Explain.
5.
Hopefully you answered “yes!” to question #4D. So now, we we should be able to create a circle
centered at point D (the incenter) that will attach to all three sides of the triangle (At points E, F
and G above). Justify this by adding the circle to your sketch. Make sure to “lock” the control
point onto point E, F or G. This circle is said to be “inscribed” in the triangle.
6. Drag the vertices of your triangle and observe that we will have an inscribed circle in all possible
triangles! By using the point of intersection of the angle bisectors (called the incenter), we can
inscribe a circle into any triangle!
PART FIVE: THE GRAND FINALE: THE EULER SEGMENT
(You will need to save the GSP sketch that you create below and email it to me when done.)
1. Create triangle XYZ.
2. Use the appropriate constructions to create a centroid. Label the centroid as C. Hide everything
other than the original triangle and the centroid.
3. Use the appropriate constructions to create an orthocenter. Label the orthocenter as O. Hide
everything other than the original triangle and the centroid and the orthocenter.
4. Use the appropriate constructions to create a circumcenter. Label the circumcenter as CC. Hide
everything other than the original triangle, the centroid, orthocenter and circumcenter.
5. Use the appropriate constructions to create an incenter. Label the incenter as I. Hide
everything other than the original triangle and the centroid, orthocenter, circumcenter and
incenter.
6. Drag your triangle to observe how the points behave. Three of the four are collinear (which
means they lie on the same line. Which three?
7. Construct a segment that contains all three collinear points. (If you are not sure what that word
means, look it up!) This is called the Euler Segment. Drag the a vertex o the triangle again and
look for interesting relationships on the Euler Segment. Document at least two observations
here. (Hint: I’d like one conclusion to address the relationship between the lengths of the two
parts of the Euler Segment.)
8.
Save your GSP file as Euler Segment – your last name. For example, if your last name is
Zuckerberg, you would save your file as Euler Segment – Zuckerberg. Then, email it to me at
leckys@mpsct.org within 3 days of our lab. Also include your answers to questions 6 and 7
immediately above in a textbox of your GSP file.