GSP CONCURRENCY LAB, CONTINUED (EXTRA CREDIT)
PART THREE: 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:
E
A
4.
D
C
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. Now,
complete the conjecture below based on our investigation. Use descriptive words without
referencing points A – E. (Hint: You’ll need the phrase “endpoints of the segment”)
Conjecture: Given a point that lies on the perpendicular bisector of a segment, …
5.
Our goal now is to prove this conjecture! 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, what we control over 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 FOUR: FINDING THE LOCATION OF A CIRCUMCENTER
If all went well in Part Three, you have successfully proved that any point on the perpendicular
bisector of a segment is equidistant from the endpoints of a segment.
Prior to this part of the lab, recall two definitions:
Circle: The set of all points equidistance from a single point called the center.
Triangle Circumcenter: The point of intersection between the three perpendicular bisectors of a
triangle.
5. Using the segment tool, create an acute triangle ABC as shown.
B
C
A
6.
Construct the three perpendicular bisectors of this triangle. (First, plot the midpoints of each
side. Then, highlight each point and its side and choose Perpendicular Line from the Construct
menu.) Carefully plot a point at the intersection of the lines. This is, of course, our
circumcenter.
7. 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
C
BONUS: JUSTIFYING THE LOCATION OF THE CIRCUMCENTER
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 the three perpendicular bisectors of a circle,
the point of intersection (the circumcenter) will indeed ALWAYS serve as the center of a circle
that passes through the three vertices of the triangle.
So, here we go!
The diagram of your construction is included again here. I have added cursive lower case letters
to name the lines:
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. So, any point on line L,, like
point D, is equidistant from the endpoints of the segment (namely points A and B). So, we
know that the length of DA is the same as the length of DB. So, we can note that
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.
Now, you take over by following through the next set of questions.
a) Similarly, what can say about point D in relation to points A and C? Justify your conclusion
as modeled above. Note the congruency that results.
b) Similarly, what can say about point D in relation to points B and C? Justify your conclusion
as modeled above. Note the congruency that results.
c) Now, write a sentence or two to justify that the intersection of the perpendicular bisectors
of a triangle will always serve as the center of a circle that passes through all vertices of the
triangle.
FYI – we say the circle formed in this way circumscribes the 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.