Geometry Summer Institute 2014
Centers of Triangles
Summary
Goals
This task provides a guided
discovery and investigation
of the points of concurrency
in triangles. Participants
will construct and use the
following points:
Participant Handouts
 Investigate different
centers of the triangle
 Discover the special
relationships of the
circumcenter,
orthocenter, incenter,
and centroid.
 Realize that three of
those points are collinear
and form the Euler Line
1. Centers of Triangles
• incenter
• orthocenter
• circumcenter
• centroid
Materials
Technology
Source
Paper
Pencil
Patty Paper
LCD Projector
Facilitator Laptop
GeoGebra
Teaching Geometry 120 minutes
in Grade 8 and High
School According to
the Common Core
Standards
H. Wu
Centroid Proofs –
Dr. Blumsack
Estimated Time
Mathematics Standards
Common Core State Standards for Mathematics
G-CO. 9, 10: Prove Geometric theorems
10. Prove theorems about triangles. Theorems include: measures of interior angles of a
triangle sum to 180; base angles of isosceles triangles are congruent; the segment joining
midpoints of two sides of a triangle is parallel to the third side and half the length; the
medians of a triangle meet at a point.
Standards for Mathematical Practice
2. Reason abstractly and quantitatively
3. Construct viable arguments and critique the reasoning of others
4. Model with mathematics
5. Use tools appropriately
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Geometry Summer Institute 2014
Instructional Plan
(Slide 2)
We now turn to the standard concurrence theorems related to a triangle. We need a
definition: three or more lines are concurrent if they meet at one point.
Participants can be divided in 4 groups and each group works on a different center or all
groups work on all centers. Whole group discussion of discovery is recommended.
Circumcenter
(Side 3)
1. Definition
i.
ii.
The perpendicular bisectors of the three sides of a triangle meet at a point,
called the circumcenter of the triangle.
There is a unique circle that passes through the vertices of a triangle, and
the center of this circle (the circumcircle of the triangle) is the
circumcenter.
2. Patty Paper investigation
i.
Draw a large acute scalene triangle on a patty paper
ii. Fold to construct the perpendicular bisector of each side of the triangle
iii. What do you notice?
iv.
Use another patty paper to find out the distance from the point of
intersection of the perpendicular bisector to each vertex of the triangle.
What do you notice?
3. GeoGebra Investigation
i.
Construct any triangle in the file
ii. Using the tool “perpendicular bisector”, construct the three perpendicular
bisectors of each side of the triangle.
iii. Place a point at the intersection (circumcenter).
iv.
Change your triangle by dragging any of the vertices of the triangle. What
do you notice? Is the circumcenter always inside of the triangle? If not,
what kind of triangle makes it go to the outside? Could the circumcenter
be on an edge of the triangle? What type of triangle is that?
v.
Find the distance from the center to each vertex of the triangle. What do
you notice? Why is this true? In other words, what geometric idea makes
this true?
vi.
Construct a circle with the center in the circumcenter that goes through a
vertex of the triangle? What do you notice? Change your triangle, are your
observations still holding true?
(Slides 4, 5)
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4. Formal Proof.
Let the triangle be ABC, as shown, and let M, N be the midpoints of BC and AC,
respectively. Also let the perpendicular bisectors of BC, AC be l1 and l2,
respectively. Let l1 and l2 meet at O.
Since O lies on the perpendicular bisector of BC, |OB| = |OC|. Similarly, |OC| =
|OA|. Together, we have |OB| = |OA| = |OC|, i.e., O is equidistant from A, B and
C. In particular, O is equidistant from A and B. O also lies on the perpendicular
bisector of AB. As O already lies on the perpendicular bisectors of BC and AC,
this proves the first part of the theorem.
To prove the second part, we have to first prove that there is a circle with center at
O that passes through the vertices A, B, and C, and that any such circle must
coincide with this circle. Consider the circle K with center O and radius |OA|. K
must pass through B and C, on account of |OB| = |OA| = |OC|, so we have proved
the existence of such a circle. Next, we have to show that any other circle K′
passing through A, B and C must coincide with K. Let the center of K′ be O′.
Since O′ is by definition equidistant from B and C, O′ lies on the perpendicular
bisector of BC and therefore lies on l1. For exactly the same reason, O′ must also
lie on l2, the perpendicular bisector of AC. Therefore O′ is the point of
intersection of l1 and l2, which is O. This shows O = O′. Since K′ passes through
A, the radius of K′ is also |OA|. Hence K′ = K because they have the same center
and the same radius.
Orthocenter
1. Definition
The three altitudes of a triangle meet at a point, called the orthocenter of the
triangle.
2. Patty Paper Investigation
i.
Draw a large acute scalene triangle on a patty paper
ii. Fold to construct the altitudes of each side of the triangle (perpendicular to
one side going through the opposite vertex)
iii. What do you notice?
3. GeoGebra Investigation
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Geometry Summer Institute 2014
i.
ii.
iii.
iv.
v.
Use the same file of the circumcenter. Hide the circumcenter, the
perpendicular bisectors, and the circumcircle.
Using the perpendicular line tool, construct the three altitudes. What do
you notice?
Construct a point at the intersection. This point is the orthocenter.
Change your original triangle. Is the orthocenter always in the inside of
the triangle? Describe in detail your observations.
When does the orthocenter fall in a triangle vertex? Why do the
perpendicular bisectors overlap the sides of the triangle?
(Slides 7, 8)
4. Formal Proof
Let the altitudes of △ABC be AD, BE and CF. Through each vertex, draw a line
parallel to the opposite side, resulting in a triangle which we denote by △A′B′C′.
If the triangle is acute, it is illustrated on the left below. However if (let us say)
∠A is obtuse, then we have a situation illustrated on the right below.
By construction, the quadrilaterals AC′BC, ABCB′ are parallelograms. Therefore,
|C′A| = |BC| = |AB′|. Thus A is the midpoint of C′B′. Moreover, since AD ⊥ BC
and BC ∥ C′B′, we also know that AD ⊥ C′B′
It follows that LAD is the perpendicular bisector of C′B′. Similarly, LFC and LBE
are perpendicular bisectors of A′B′ and C′A′, respectively. Then, LAD, LFC and
LBE meet at the circumcenter of △A′B′C′
Incenter
1. Definition
The three angle bisectors of a triangle meet at a point, called the incenter of the
triangle. The incenter is the unique point equidistant from the three sides.
2. Patty Paper Investigation
i.
Draw a large acute scalene triangle on a patty paper
ii. Fold to construct the angle bisectors of each angle of the triangle
iii. What do you notice?
iv.
Compare the distances from the incenter to the sides of the triangle. Also
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Geometry Summer Institute 2014
compare the distances from the incenter to the vertices of the triangle?
What did you discover about the incenter of a triangle? What distances
seem to be the same? Write your conjecture.
3. GeoGebra Investigation
i.
Use the same file, hide all other lines leaving the triangle.
ii. Using the angle bisector tool, construct the three angle bisectors. Construct
the point of intersection, called the incenter.
iii. Change the triangle and observe the incenter. Does the incenter stay
always in the inside of the triangle? Describe your observations in detail
iv.
Measure the distance from the incenter to the sides of the triangle and also
measure the distance from the incenter to the vertices of the triangle. What
distances seem to be the same? Change your triangle and observe the
distances? What do you conclude?
v.
Construct a line perpendicular to one side of the triangle that goes through
the incenter. Draw a point at the intersection of the perpendicular line and
the side.
vi.
Now construct a circle with center at the incenter that goes to the point of
intersection between the perpendicular line and the side.
vii.
Change your triangle. What do you notice?
4. Formal Proof
Let the angle bisectors AE and BD of ∠A and ∠B in △ABC, respectively,
intersect at I.
Since I lies on the angle bisector of ∠A, it is equidistant from AC and AB.
Because I also lies on the angle bisector of ∠B, it is equidistant from BA and BC.
Together, these two facts imply that I is equidistant from CA and CB. I must also
lie on the angle bisector of ∠C. So all three-angle bisectors are concurrent. Now
suppose there is another point I′ equidistant from all three sides. Because I′ is
equidistant from AB and AC, it implies that I′ lies on the angle bisector of ∠A.
The same reasoning then shows that I′ lies on all the angle bisectors, i.e., I′ = I.
Centroid
(Slide 11)
1. Definition
The three medians of a triangle meet at a point G, called the centroid of the
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triangle.
2. Patty Paper Investigation
i.
Draw a large acute scalene triangle on a patty paper
ii. Locate the midpoint of each side of the triangle and fold or draw the
medians. What do you notice?
iii. Will your observations hold true for any type of triangle? Write a
conjecture about the medians of the triangle
3. GeoGebra Investigation
i.
Use the same file. Hide all lines, points, and circle.
ii. Using the midpoint tool, construct the midpoint of each side of the triangle
iii. Using the segment tool, draw the three medians. What do you observe?
iv.
Draw a point at the intersection of the medians. This is the centroid.
v.
Change your triangle. Does the centroid always stay in the inside? Write
your conjecture.
vi.
Using only one median, find the distance between the midpoint of the side
to the centroid and from the centroid to the connecting vertex. What do
you observe? Move the triangle, is your conjecture holding true? Use
another median and do the same measurement. Write a conjecture about
the centroid.
4. Formal Proof
(Slides 12 to 20)
This proof assumes that we have already proven that the diagonals of a
parallelogram bisect each other and that in a triangle the segment connecting the
midpoints of two sides is parallel to the third side of the triangle.
We focus attention on the median issuing from B, to be called BB′. We claim that
either of the two medians issuing from A and C will meet BB′ at a point G so that
|BG| = 2|GB′|. Once this is done, then we know that all three medians meet at the
point G as described and the theorem will be proved.
Let us consider the case of the median CC′ issuing from C. We will prove
something that is equivalent to the preceding assertion, namely, we let G be the
point of intersection of CC′ and BB′ and then prove that |BG| = 2|GB′|. We know
that C′B′ ∥ BC and |C′B′| = ½ |BC|. Now let M, N be midpoints of BG and CG,
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Geometry Summer Institute 2014
respectively. Then, MN ∥ BC and |MN| =1/2 |BC|. Therefore, C′B′ ∥ MN and
|C′B′| = |MN|. MNB′C′ is a parallelogram and therefore the diagonals MB′ and
NC′ bisect each other. Thus, |GB′| = |MG|, but since M is the midpoint of BG, we
have |BM| = |MG| = |GB′|, which is equivalent to |BG| = 2|GB′|.
Another proof using areas of triangles
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Given triangle ABC.
Construct TWO medians BE and CD.
Indicate F as the intersection of these TWO medians.
Construct ray through A & F, intersecting BC at point G.
Let A1, … A6 denote the areas of the six triangles.
Denote p=CG and q=BG
Objective: prove that AG is the THIRD median, i.e., p=q
A1=A6 and A2=A3
Reason: equal base lengths and heights.
A1+A2+A3=A6+A1+A2
Reason: each equals half the area of triangle ABC since E and D being
midpoints.
Therefore, A3=A6 and A1=A2=A3=A6
Let T be this common value
A4/A5=p/q and q(A4)=p(A5).
Reason: Triangles CFG and BFG have the same altitudes
(A2+A3+A4)/(A1+A5+A6)=p/q => q(2T+A4)=p(2T+A5)
Reason: Triangles AGC and AGB have the same altitudes
2Tq + q(A4) = 2Tp + p(A5) =>p=q
Proof using Coordinates
Conjecture: The medians of a triangle ABC intersect at a point that is 2/3 of the
way from each vertex to the midpoint of the opposite side.
Let the vertices of triangle ABC be represented as
A: (a1,a2), B: (b1,b2), C: (c1,c2)
1. Determine the coordinates of the midpoint (G) of side BC.
Answer: ((b1+c1)/2, (b2+c2)/2)
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2. Determine the coordinates (x, y) of the point that is 2/3 of the way
from A to G.
Answer: x= (1/3)a1 + (2/3)(1/2)(b1+c1)
(x, y) = ( (1/3)(a1+b1+c1), (1/3)(a2+b2+c2) )
3. What is the corresponding result for the other medians? Explain.
Answer: All that we need do is permute the symbols a, b, c. The
result is the same point (x, y).
Euler Line
The Euler line, named after Leonhard Euler, is a line determined from any triangle that is
not equilateral. It passes through several important points determined from the triangle.
Using your GeoGebra file, determine what points are collinear and determine the Euler
line.
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