Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Geometry, Module 1, Lesson 19 – Using Logical Reasoning to Prove Conjectures About
Circles
Part I: Lesson title and student performance descriptor and Key Words
Title: Using Logical Reasoning to Prove Conjectures About Circles
Student performance descriptor: Given conjectures about circles, the student will use
deductive reasoning and counter examples to prove or disprove the conjectures.
Key Words:
Central angle
Inscribed angle
Arc
Intercepted arc
Semicircle
Chord
Part II: TEKS addressed
(G.3) Geometric structure. The student applies logical reasoning to justify and prove mathematical statements. The
student is expected to:
(C) use logical reasoning to prove statements are true and find counter examples to disprove statements that are
false.
Part III: Four to five TAKS-like assessment items.
Question 1
Source: Author on Sketchpad
1
Linda Shaub
lshaub@ipsi.utexas.edu
Answer Choice
Deb Erhart
Correct Answer Feedback
Incorrect Answer Feedback
̂ ≅ 𝐴𝐶
̂
A. 𝐴𝐵
Incorrect. Although they
appear to be approximately the
same length in the diagram,
there is no information to
guarantee that relationship.
̅̅̅̅ is parallel to 𝐵𝐷
̅̅̅̅.
B. 𝐴𝐹
Incorrect. There is not enough
information to conclude this
parallel situation
C. ∆𝐵𝐸𝐷 is isosceles, so ∠𝐵 ≅
∠𝐷, since base angles of
isosceles triangles are
̂ ≅ 𝐶𝐷
̂.
congruent 𝐴𝐵
Incorrect. ∆𝐵𝐸𝐷 is not
necessarily an isosceles
triangle, so the congruence
conclusions cannot be made.
D. This situation is not
possible. The two inscribed
angles would have to have
equal measures.
Correct! Inscribed angles
sharing the same intercepted
arc are congruent, so ∠𝐶 must
be congruent to ∠𝐷.
Question 2
Choose the statement that is true under all conditions.
Answer Choice
Correct Answer Feedback
Incorrect Answer Feedback
A. Congruent chords in the
same circle must be parallel
and equidistant from the
center.
Incorrect. The chords must be
equidistant from the center, but
not necessarily parallel.
B. Two equilateral triangles are
formed from congruent
chords of a circle and the
radii including the endpoints
of these chords.
Incorrect. This is possible only
if the chords are the same
length as the radii of the circle.
C. Two isosceles triangles are
formed from congruent
chords of a circle and the
radii including the endpoints
of these chords.
Correct! They are congruent
because of SSS Postulate. The
radii are all congruent and the
congruent chords form the third
congruent correspondence.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
D. Any chord perpendicular to
another chord in a circle is
the perpendicular bisector of
that chord.
Incorrect. This is only possible
when the chord is the
diameter.
Question 3
Choose the statement that is true under all conditions.
Answer Choice
Correct Answer Feedback
Incorrect Answer Feedback
A. Parallel chords in a circle
must be congruent.
B. Two parallel chords in a
circle intercept congruent
arcs.
Incorrect. This would happen
only if the chords were
equidistant from the center.
Correct! You can prove this by
drawing a transversal, showing
alternate interior angles (which are
also inscribed angles) are
congruent, then the intercepted
arcs being congruent.
Incorrect. This would only
happen if the intercepted arcs
are each 90 degrees.
C. Two parallel chords in a
circle intercept
supplementary arcs.
D. One chord of a pair of
parallel chords in a circle
must contain the center
of the circle.
Incorrect. This means that the
diameter must be one of the
parallel chords. This is only
one possibility.
Question 4
What determines a perpendicular bisector of a chord in a circle?
Answer Choice
A. The perpendicular must
go through the center of
the circle to the chord.
Correct Answer Feedback
Correct! By going through the
center of the circle, the chord
is bisected and the angle
formed is 90 degrees.
3
Incorrect Answer Feedback
Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Incorrect. An infinite number
of chords could go through the
midpoint of the chord. Only
one will be perpendicular.
B. The chord must go
through the midpoint of
the chord.
Incorrect. Many chords will
bisect the chord, but only one
will be perpendicular.
C. The chord must bisect
the chord.
Incorrect. Many chords will
intersect at a 90-degree
angle, but only one also goes
through the midpoint.
D. The chord must form a
right angle to the chord.
Part IV: Identify the sections of the lesson.
Section 1: Some Conjectures Using Inscribed Angles
Section 2: Some Conjectures About Parallel Chords in a Circle
Section 3: Some Conjectures About Congruent Chords in a Circle
Part V: The lesson as it will appear to the students in Epsilen
Page 1
Section 1: Some Conjectures Using Inscribed Angles
Note to developer: Please show the definition of the vocabulary highlighted in yellow.
There are two very important angle and arc relationships when working with circles. The first
one is about central angles and their intercepted arcs.
Source: author, created on Sketchpad.
Alt=“Circle A with central angle CAB.”
The second relationship is about inscribed angles and their intercepted arcs.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Alt=“Circle A with inscribed angle BCD.”
Source: author, created on Sketchpad.
To see a demonstration of the relationship between the central and inscribed angles of
the same intercepted arc, click on the picture below. Be sure to drag the vertices to
compare angle CAB to angle CDB.
Note to developer: I wasn’t able to get the applet to work correctly but the writer can….. (If you
are unable to get the applet to move as stated above would it be possible to create an
interactive where the student is able to move each one of the points A, B, or C at a time around
the circle to see the different angles? Thanks! If not please let me know.) Please embed the
applet- http://demonstrations.wolfram.com/InscribedAndCentralAnglesInACircle/.
Source: Wolfram Demonstrations Project, Inscribed and Central Angles in a Circle, Jay Warendorff
We are going to use these relationships to investigate some conjectures about circles. Then we
will use deductive reasoning to see if our conjectures hold true in all situations.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Suppose you have an inscribed angle whose sides go through the endpoints of a diameter. In
other words, the inscribed angle is included in a semicircle. What conjecture can you make
about this angle? Let’s investigate.
Definition needed - Semicircle: Half a circle
Example 1
Example 2
Look at example 1. What do you notice?
Look at example 2. Does the relationship you
noticed in example 1 hold for angles E and F?
(Hint: Use the corner of a piece of paper to
check.)
Source: author, created on Sketchpad
Alt=“Circle A with one inscribed angle in a
semicircle.”
Alt=“Circle A w/ 2 added inscribed angles in
semicircle”
In the “Take Notes” section or on your own piece of paper, complete the following conjecture
then click on the blanks to check your answer:
Note to developer: When the student clicks on the blank have the answers appear and stay
visible. The answers are highlighted in green.
Inscribed angles in a _semicircle _ are always _right__ angles.
Here you are really just guessing. How could you know for sure that your conjecture is always
true?
We can show this by using deductive reasoning.
We know that inscribed angles are half their intercepted arc. How could you use this information
to prove that these inscribed angles with sides going through the endpoints of a diameter (in a
semicircle) will always be right angles? Write your response in your Take Notes tool.
Click here to compare your response.
Please create a popup when they click this link with the following answer:
The intercepted arc of a semicircle is 180 degrees. Therefore, any inscribed angle contained in a semicircle
will have an intercepted arc of 180 degrees. The angle will be half that, making its measure 90 degrees. Any
90 degree angle is a right angle.
6
Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Note that all the inscribed angles had the same intercepted arc in the examples above.
However, inscribed angles do not always have sides that go through the endpoints of a
diameter.
Let’s look at another situation where we have inscribed angles that share the same intercepted
arc, but are not necessarily right angles. Click on the picture below and be sure to drag points to
compare the two angles that share the same intercept.
Note to developer: (I wasn’t able to get the applet to work as stated by the writer. Checked with
the writer, it works for them. If it doesn’t work would you please create an applet that does what
is stated above? If not please let me know. Thank you! – The student is able to move one point
at a time around the circle creating different angles.) Please embed the applet http://demonstrations.wolfram.com/InscribedAnglesThatInterceptTheSameArc/
Source: Wolfram Demonstrations Project, Inscribed Angles That Intercept
The Same Arc, Jay Warendorff
In the “Take Notes” section or on your own piece of paper, complete the following conjecture
then click on the blank to check your answer:
Inscribed angles sharing the same intercepted arc are __congruent____.
Note to developer: When the student clicks on the blank the answer appears and stays visible.
The answer is highlighted in green.
The demonstration above gives you an idea that these angles will always be congruent. To be
̂)
sure, we can prove this deductively. (NOTE: arc 𝐴𝐵 is 𝐴𝐵
𝒎∠𝑪 =
𝟏
̂
𝒎𝑨𝑩
𝟐
𝒎∠𝑫 =
𝟏
̂
𝒎𝑨𝑩
𝟐
∴ 𝒎∠𝑪 = 𝒎∠𝑫
If angles have the same measures, then they are congruent, so C  D.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
How could you use the same reasoning to prove the A  B?
Click here to compare your answer.
Please create a popup when they “click here” showing the following answer:
̂ to 𝐶𝐷
̂.
You could use the same format, except change angles C and D to A and B, and 𝐴𝐵
In the next section, we prove some conjectures about chords of a circle.
Section 2: Some Conjectures About Parallel Chords in a Circle
Page 2
Now we will investigate and prove some conjectures about chords in a circle. Suppose we have
a circle containing 2 parallel chords. Click on the protractor below and look for relationships.
What do you notice about the intercepted arcs?
Note to developer: This will be a demonstration where the protractor is the same size as half the circle.
Student will click on the protractor and it will move over to the circle with the center of the protractor
lined up with the center of the circle (Point A – Figure 1 – a rough sketch). Rotate the protractor until the
“0” falls on Point E – Figure 2 – Rough sketch. Have Point B flash to capture student’s attention, then
the red dashed line appears. A few seconds later, have the protractor rotate to the other side of the
circle with the “0” on the right lining up with Point D. Have Point C flash – Figure 3 sketch – then the red
dashed line appears.
The blue is an imitation protractor
used to show the examples. Please
use the protractor in the
interactive.
Source: Mathisfun.com, Measuring Degrees
Source: Author created on Sketchpad
Figure 1
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Deb Erhart
0
180
Linda Shaub
lshaub@ipsi.utexas.edu
Protractor replica
Figure 2
Figure 3
Copy the following in the “Take Notes” section or on your own piece of paper filling in the
blanks.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
The measure of the first arc is ____ and the measure of the second arc is ____.
Complete the following conjecture: If a circle contains parallel chords, those chords intercept
_____congruent ___ arcs.
Note to developer: When the student clicks on the blank have the answers appear and stay
visible. The answer is highlighted in green.
Now let’s prove the conjecture.
Complete the following drag and drop puzzle to prove that this conjecture is true. Use the
pictures to help you organize the statements and reasons in your proof.
Note to developer: This is a drag and drop puzzle. When the student moves to the incorrect
position, I want it to pop back to the menu. The items to place are in the table below. The
answers are in green in the key.
𝑚 𝐸  𝑚 𝐶
the measures of inscribed
angles are half their
intercepted arcs.
Circle A with
parallel chords BC
and ED.
1
1
̂
̂ ,  𝐸  𝑚𝐶𝐷
𝑚∠C= 2 𝑚𝐸𝐵
2
Draw segment
EC.
Congruent
arcs have
equal
measures.
1
1
̂ = 𝑚𝐸𝐵
̂
𝑚CD
2
2
Alternate interior
angles of parallel
lines cut by a
transversal are
congruent.
̂ = 𝑚𝐸𝐵
̂
𝑚𝐶𝐷
̂ ≅ 𝐸𝐵
̂
∴ 𝐶𝐷
Substitution
Multiplicative
Property of
Equality
Given: Circle A with parallel chords BC and ED
Draw segment EC
Two points determine a line, so segments may be drawn.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
𝑚 𝐸  𝑚 𝐶, because alternate interior angles of parallel
lines cut by a transversal are congruent.
1
1
̂ because the measures of
̂ , 𝑚 𝐸  𝑚𝐶𝐷
𝑚∠C= 2 𝑚𝐸𝐵
2
inscribed angles are half their intercepted arcs.
1
̂ = 1 𝑚𝐸𝐵
̂
𝑚CD
2
2
because of substitution.
because of the multiplicative property of equality.
̂ = 𝑚𝐸𝐵
̂
𝑚𝐶𝐷
̂ ≅ 𝐸𝐵
̂ because congruent arcs have equal measures.
∴ 𝐶𝐷
What about the converse of your conjecture that you have proved? Do you think the converse
would hold true?
Write the converse in the “Take Notes” section.
Click here to compare your answer.
Please create a popup when they click this link with the following answer:
If a circle’s chords intercept congruent arcs, then they are parallel.
Is this statement always true? On notebook paper, try to draw a situation where this statement
would be false. If you can, then you have shown a counter example to prove that this statement
is not always true. Did you find one?
Look at the example given below.
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Linda Shaub
lshaub@ipsi.utexas.edu
Source: Author created in Sketchpad
Deb Erhart
Alt=“Circle with two congruent, nonparallel chords.”
It appears to be possible to find congruent arcs from nonparallel chords. Therefore the converse
is not true.
The next section will investigate the relationship of two congruent chords in a circle that are not
necessarily parallel.
Section 3: Some Conjectures About Congruent Chords in a Circle
Watch the following demonstration as the chords are placed in the circle. Click on the circle to
begin the animation. What do you notice?
Note to developer: When the student clicks on the circle, have each chord move one at a time to
the circle. The shortest chord will be the farthest from the center. The key is given below the
activity.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
KEY
In the “Take Notes” section or on your own piece of paper, record what you noticed about the
position of the chords of different lengths.
Click here to compare your answer.
Please create a popup when they click this link with the following answer:
The longest chord goes through the center. The chords are shorter the farther away from the center they are.
What is the name of the longest chord?
Click here to check your answer.
Please create a popup when they click this link with the following answer:
diameter
Is it possible to have a chord longer than the diameter? Why or why not?
Click here to compare your answer.
Please create a popup when they click this link with the following answer:
No, any segment longer that the diameter would have at least one endpoint outside the circle.
So thinking about distance from the center, make a conjecture about 2 chords that would be the
same length in a circle.
In the “Take Notes” section or on your own piece of paper, complete the following conjecture:
Two chords in the same circle are congruent if they are ____ equal length from the center
_______. Check answer.
Note to developer: When the student clicks “Check answer” have the
answers appear and stay visible. The answer is highlighted in green above.
13
Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Let’s prove it. We can have 2 congruent chords that are not necessarily parallel.
Let CD and AB be congruent chords on the same
CD  AB
circle.




Alt=“circle
with 2
nonparallel
chords”
Source: Author on Sketchpad
Note to developer: This is a drag and drop puzzle. When the student moves to the incorrect
position, I want it to pop back to the menu. The items to place are in the table below. The
answers are in green in the key.
̅̅̅̅ , 𝐷𝐸
̅̅̅̅ and ̅̅̅̅
̅̅̅̅, 𝐴𝐸
We can draw 𝐶𝐸
𝐵𝐸 . Complete the following drag and drop puzzle to make a
deductive argument to show congruent chords in the same circle are equidistant from the center
of that circle.
Isosceles
radii
SSS Postulate -
(If three sides of
one triangle are
congruent to
three sides of
another triangle,
then the triangles
are congruent)
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circle
congruent
Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
̅̅̅̅ , 𝐷𝐸
̅̅̅̅ and ̅̅̅̅
̅̅̅̅, 𝐴𝐸
𝐶𝐸
𝐵𝐸 are the radii of the __ circle __.
Complete the following:
The _radii ____of the same circle are congruent.
__ Isosceles __triangles are formed with these radii and
congruent chords.
Source: Author on Sketchpad
These triangles are _ congruent __ because of the
__ SSS Postulate __.
Alt=”2 triangles in circle”
When the student “Clicks Here” the red line on the left appears then the one on the right appears.
Click Here
Alt=”2 triangles in circle”
Source: Author on Sketchpad
Altitudes ̅̅̅̅
𝐸𝐺 and ̅̅̅̅
𝐸𝐹 are formed from Point 𝐸, the center of the circle, to ̅̅̅̅
𝐶𝐷 and ̅̅̅̅
𝐴𝐵.
Remember that altitudes form right angles. The logical argument below proves that the altitudes
are congruent. Click each step to view the proof.
Step 1
When the student clicks “Step 1” the information in green appears
Remember that altitudes of isosceles triangles are also medians. Therefore they cut the base
segment into 2 congruent pieces. 𝐺 and 𝐹 are midpoints.
1
1
̅̅̅̅ ≅ 𝐶𝐷
̅̅̅̅ and 𝐹𝐵
̅̅̅̅.
̅̅̅̅ ≅ 𝐴𝐵
Therefore, 𝐺𝐷
2
2
1
̅̅̅̅ are congruent, then 1 ̅̅̅̅
Since our original chords ̅̅̅̅
𝐶𝐷 and 𝐴𝐵
𝐶𝐷 ≅ 2 ̅̅̅̅
𝐴𝐵 and ̅̅̅̅
𝐺𝐷 ≅ ̅̅̅̅
𝐹𝐵 by
2
substitution.
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Step 2
Our radii are congruent also, so we now have congruent triangles
EGD  EFB
by HL (Hypotenuse-Leg Theorem - If two triangles have congruent hypotenuses and a
corresponding leg congruent, then the triangles are congruent.).
So EG  EF
By CPCTC (Corresponding Parts of Congruent Triangles are Congruent).

 3
Step
Since the altitudes are congruent, then the chords CD and
AB are equidistant from the center.


Source: Author on Sketchpad
Alt= “2 triangles w/ altitudes and congruencies marked”
Looking at the diagram above, note that EG and EF both form 90-degree angles and cut the
segments in half. This makes them not only medians and altitudes, but also ______
perpendicular bisectors ______. Check answer.
Note to developer:
When thestudent clicks “Check answer” have the
answers appear and stay visible. The answer is highlighted in green above.
Note that the perpendicular bisectors pass through the center of the circle.
There is more information we can prove from congruent chords. Remember we showed that
when we include the radii from the endpoints of the chords, we have isosceles triangles.
Complete the following drag and drop puzzle to discover another congruent chord relationship.
Note to developer: This is a drag and drop puzzle. When the student moves to the incorrect
position, I want it to pop back to the menu. The items to place are in the table below. The
answers are in green in the key.
16
Linda Shaub
lshaub@ipsi.utexas.edu
Central
substitution CD
Deb Erhart
AB
CPCTC
Intercepted
SSS
CED  AEB by SSS Postulate.
So  CED   AEB by ___ CPCTC ____.
m CED = m Arc __ CD __and
m AEB = m Arc ____ AB ________,
because the measure of __ central ___ angles
equals the measures of their _ intercepted __
arcs.
̂ = 𝑚𝐴𝐵
̂ , because of _ substitution __.
So 𝑚𝐶𝐷
̂ ≅ 𝐴𝐵
̂
 𝐶𝐷
Source: Author by Sketchpad
Alt: “2 triangles w/ congruent radii and
chords marked”
In other words, congruent chords have ___ congruent arcs ________. Check answer.
Note to developer: When the student clicks “Check answer” have the
answers appear and stay visible. The answer is highlighted in green above.
You have given deductive arguments for conjectures about chords in circles. If you would like
more practice, look at all the converse statements to see if they hold true. Remember to look for
counterexamples to prove a statement is not true in all cases.
Additional Resources
Reading Protractors - How to read a protractor
http://www.mathsisfun.com/geometry/protractor-using.html
Circle Vocabulary - A review of circle vocabulary
http://www.mathsisfun.com/geometry/circle.html
Circle Theorems and Angle Relationships - A review of circle theorems and angle relationships
http://www.mathsisfun.com/geometry/circle-theorems.html
Parts of a Circle - Interactive examples of parts of a circle
http://www.mathopenref.com/circle.html
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Linda Shaub
lshaub@ipsi.utexas.edu
Deb Erhart
Perpendicular Bisector of a Chord Conjecture – The website provides an explanation of the
perpendicular bisector of a chord.
http://www.geom.uiuc.edu/~dwiggins/conj41.html
Inscribed Angle Conjecture – The website provides an explanation of the inscribed angle
conjecture.
http://www.geom.uiuc.edu/~dwiggins/conj44.html
Resources Used in This Lesson
Central and Inscribed Angles – The website provides a demonstration of the relationship
between the central and inscribed angles of the same intercepted arc.
http://demonstrations.wolfram.com/InscribedAndCentralAnglesInACircle/
Inscribed Angles and Their Intercepted Arcs - A demonstration is provided for the relationship of
inscribed angles sharing the same intercepted arc.
http://demonstrations.wolfram.com/InscribedAnglesThatInterceptTheSameArc/
Protractor – A copy of a protractor.
http://www.mathsisfun.com/geometry/degrees.html
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