options: Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints.

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.

Prove theorems about parallelograms. Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals.
problem_activity: Activity
The proof that if a point
\(C\)
is the same distance from
\(A\)
as it is from
\(B\)
, then
\(C\)
must be on the perpendicular bisector of
\(AB\)
is challenging. Students may struggle to make sense of what it means to prove that a point must be on a line. This proof, like the proof of the Isosceles Triangle Theorem, requires thinking carefully about how to define an auxiliary line.
Making dynamic geometry software available gives students an opportunity to choose appropriate tools strategically (MP5).
Launch
Arrange students in groups of 2. Use the Notice and Wonder instructional routine with this image to launch the activity.
Display the image for all to see. Ask student to think of at least one thing they notice and at least one thing they wonder. Give students 1 minute of quiet think time, and then 1 minute to discuss the things they notice with their partner, followed by a whole-class discussion.
\(\overline{AP} \cong \overline{BP}\)
[IMG]
Things students may notice:
It’s a triangle.
Segment
\(PA\)
is congruent to segment
\(PB\)
.
Angle
\(A\)
is congruent to angle
\(B\)
(by the Isosceles Triangle Theorem).
There’s a dotted line.
Things students may wonder:
Is the dotted line the line of symmetry?
Is the dotted line the angle bisector of angle
\(APB\)
?
Is the dotted line the perpendicular bisector of segment
\(AB\)
?
The purpose of this launch is to elicit the idea that special lines through
\(P\)
when segments
\(PA\)
and
\(PB\)
are congruent have many overlapping properties, which will be useful when students draw auxiliary lines later in this activity. While students may notice and wonder many things about these images, wondering about the line through
\(P\)
is an important focus of the discussion.
If students have access to GeoGebra Geometry from Math Tools, suggest that it might be a helpful tool in this activity.
Reading, Listening, Conversing: MLR6 Three Reads.
Use a modified version of this routine to support reading comprehension of this problem. Use the first read to orient students to the situation. Ask students to describe what the situation is about (Diego, Jada, and Noah want to prove that if a point
\(C\)
is the same distance from
\(A\)
as it is from
\(B\)
, then
\(C\)
must be on the perpendicular bisector of
\(AB\)
.). Use the second read to identify geometric relationships stated by Diego, Jada, or Noah. Listen for and amplify important relationships such as “In Noah’s approach,
\(CD\)
is the angle bisector of angle
\(ACB\)
” and “In Jada’s approach,
\(D\)
is the midpoint of
\(AB\)
.” After the third read, ask students to brainstorm possible questions to help Jada and Noah prove that
\(CD\)
must be the perpendicular bisector of
\(AB\)
. This routine will help students make sense of the language in the word problem and the reasoning needed to solve the problem.
Design Principle(s): Support sense-making
Representation: Internalize Comprehension.
Provide appropriate reading accommodations and supports to ensure students access to written directions, word problems, and other text-based content.
Supports accessibility
for:
Language; Conceptual processing
Student Facing
Diego, Jada, and Noah were given the following task:
Prove that if a point
\(C\)
is the same distance from
\(A\)
as it is from
\(B\)
, then
\(C\)
must be on the perpendicular bisector of
\(AB\)
.
At first they were really stuck. Noah asked, “How do you prove a point is on a line?” Their teacher gave them the hint, “Another way to think about it is to draw a line that you know
\(C\)
is on, and prove that line has to be the perpendicular bisector.”
They each drew a line and thought about their pictures. Here are their rough drafts.
Diego’s approach: “I drew a line through
\(C\)
that was perpendicular to
\(AB\)
and through the midpoint of
\(AB\)
. That line is the perpendicular bisector of
\(AB\)
and
\(C\)
is on it, so that proves
\(C\)
is on the perpendicular bisector.”
[IMG]
Jada’s approach: “I thought the line through
\(C\)
would probably go through the midpoint of
\(AB\)
so I drew that and labeled the midpoint
\(D\)
. Triangle
\(ACB\)
is isosceles, so angles
\(A\)
and
\(B\)
are congruent, and
\(AC\)
and
\(BC\)
are congruent. And
\(AD\)
and
\(DB\)
are congruent because
\(D\)
is a midpoint. That made two congruent triangles by the Side-Angle-Side Triangle Congruence Theorem. So I know angle
\(ADC\)
and angle
\(BDC\)
are congruent, but I still don’t know if
\(DC\)
is the perpendicular bisector of
\(AB\)
.”
[IMG]
[IMG]
Noah’s approach: “In the Isosceles Triangle Theorem proof, Mai and Kiran drew an angle bisector in their isosceles triangle, so I’ll try that. I’ll draw the angle bisector of angle
\(ACB\)
. The point where the angle bisector hits
\(AB\)
will be
\(D\)
. So triangles
\(ACD\)
and
\(BCD\)
are congruent, which means
\(AD\)
and
\(BD\)
are congruent, so
\(D\)
is a midpoint and
\(CD\)
is the perpendicular bisector.”
With your partner, discuss each student’s approach.
What do you notice that this student understands about the problem?
What question would you ask them to help them move forward?
Using the ideas you heard and the ways you think each student could make their explanation better, write your own explanation for why
\(C\)
must be on the perpendicular bisector of
\(A\)
and
\(B\)
.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Student Response.
Student Facing
Are you ready for more?
Elena has another approach: “I drew the line of reflection. If you reflect across
\(C\)
, then
\(A\)
and
\(B\)
will switch places, meaning
\(A'\)
coincides with
\(B\)
, and
\(B'\)
coincides with
\(A\)
.
\(C\)
will stay in its place, so the triangles will be congruent.”
1. What feedback would you give Elena?
2. Write your own explanation based on Elena‘s idea.
Student Response
Teachers with a valid work email address can
click here to register or sign in
for free access to Extension Student Response.
Anticipated Misconceptions
Remind students who struggle with their critique or edits to make use of the tips from the display, such as asking for 3 statements and 3 reasons, or looking for congruent triangles.
Activity Synthesis
Focus discussion on what students proved and the implications. Display the image from the launch of this activity, and ask what must be true about this image, based on what they just proved. (The dotted line is the perpendicular bisector of
\(AB\)
. The dotted line is the line of reflection of
\(AB\)
.)
\(\overline{AP} \cong \overline{BP}\)
[IMG]
Display this image of the construction of a perpendicular bisector and ask students how what they just proved explains why this construction works. (Both circles have radius
\(AB\)
, so the intersection points of the circles are the same distance from
\(A\)
and
\(B\)
. Therefore, both intersection points are on the perpendicular bisector of
\(AB\)
.)
[IMG]
Ask students to add part 1 of the Perpendicular Bisector Theorem to their reference charts as you add it to the class reference chart:
If a point
\(C\)
is the same distance from
\(A\)
as it is from
\(B\)
,
\(C\)
must be on the
perpendicular bisector
of
\(AB\)
. (Theorem)
\(\overline{AC} \cong \overline{BC}\)
, so
\(C\)
is on the line through midpoint
\(M\)
perpendicular to
\(\overline{AB}\)
[IMG]


correct_option_index: 0