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Illustrative
Mathematics
G-CO Congruent angles in
isosceles triangles
Alignments to Content Standards: G-CO.C.10
Task
Below is an isosceles triangle
ABC with |AB| = |AC|:
Three students propose different arguments for why m(∠B)
= m(∠C).
a. Ravi says
If I draw the bisector of
∠A then this is a line of symmetry for △ABC and so
m(∠B) = m(∠C).
b. Brittney says
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Illustrative
Mathematics
If M is the midpoint of
⎯⎯⎯⎯⎯⎯⎯⎯
BC then △ABM is congruent to △ACM and so ∠B and ∠C
are congruent.
c. Courtney says
If P is a point on
←→
⎯⎯⎯⎯⎯⎯⎯⎯
BC such that AP is perpendicular to BC then △ABP is congruent to
⎯⎯⎯⎯⎯⎯⎯⎯
△ACP and so ∠B and ∠C are congruent.
Fill in the details in each argument to show why m(∠B)
different argument showing that m(∠B) = m(∠C)?
= m(∠C). Can you find another
IM Commentary
The goal of this task is to establish that base angles in an isosceles triangle are
congruent. There are many approaches to the task, several of which are presented
here. There are also many possible approaches to this task in the classroom:
For a more directed approach, the teacher may wish to suggest one method or
perhaps give different ideas to different groups of students.
Students might be prompted without guidance to prove that base angles in an
isosceles triangle are congruent. As they progress, the teacher can share some of the
approaches given here.
Students might discuss the relative merits of the different arguments. Which is
easiest to follow? Which assumes the least background knowledge?
Two additional arguments are presented below, one very similar to Ravi's but using
triangle congruence instead of reflections and the other similar to Courtney's argument
but using trigonometric ratios instead of triangle congruence.
Edit this solution
Solution
a. Pictured below is the line ℓ which bisects angle A:
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Illustrative
Mathematics
Since ℓ contains A, reflection about ℓ maps
A to itself. Since m(∠DAB) = m(∠DAC)
−→ −→
−
we know that reflection about ℓ maps AB to AC and vice versa. Finally, reflections
preserve distance as well as angles so this means that B and C are interchanged by this
reflection. Since reflection over ℓ interchanges angles B and C we see that
m(∠B) = m(∠C).
⎯⎯⎯⎯⎯⎯⎯⎯
b. The midpoint M of BC is pictured below:
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Illustrative
Mathematics
⎯⎯⎯⎯⎯⎯⎯⎯
Since M is the midpoint of BC this means that |BM| = |CM|. We know by hypothesis
that |AB| = |AC|. Since |AM| = |AM| we can conclude by SSS that △ABM is congruent
to △ACM. Since ∠B and ∠C are corresponding parts of congruent triangles they are
congruent.
c. Suppose P is the point on
←→
⎯⎯⎯⎯⎯⎯⎯⎯
BC so that AP is perpendicular to BC as pictured below:
⎯⎯⎯⎯⎯⎯⎯⎯
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Illustrative
Mathematics
By hypothesis ∠BPA and ∠CPA are right angles. We know that |AC| = |BC| by
hypothesis and |AP| = |AP|. By the Pythagorean theorem, |BP| = |CP|, and so by SSS
△BPA is congruent to △CPA. Hence m(∠B) = m(∠C).
One alternate proof, building on the picture from Courtney's solution, uses
trigonometric ratios. Using the picture above, we have sin ∠B = |AP| and
sin ∠C =
|AP|
. Since
|AC|
|AB|
|AB| = |AC| by hypothesis this means that sin ∠B = sin ∠C.
Since these are both acute angles, this is only possible if m(∠B) = m(∠C).
A second alternate proof builds on Ravi's picture. We have |AC| = |AB| by hypothesis
and |AD| = |AD|. We also know that m(∠BAD) = m(∠CAD) by hypothesis. By SAS,
△BAD is congruent to △CAD and we again conclude that ∠B is congruent to ∠C.
G-CO Congruent angles in isosceles triangles
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