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Illustrative
Mathematics
G-SRT Joining two midpoints of
sides of a triangle
Alignments to Content Standards: G-SRT.B.4
Task
⎯⎯⎯⎯⎯⎯⎯
Suppose ABC is a triangle. Let M be the midpoint of AB and ℓ the line through M
←→
parallel to AC :
a. Show that angle CAB is congruent to angle PMB and that angle BPM is congruent
to angle BCA. Conclude that triangle MBP is similar to triangle ABC.
⎯⎯⎯⎯⎯⎯⎯⎯
b. Use part (a) to show that P is the midpoint of BC .
IM Commentary
This task is closely related to very important material about similarity and ratios in
geometry. In this case, the point M selected on the triangle
More generally, consider the picture below:
⎯⎯⎯⎯⎯⎯⎯
ABC is the midpoint of AB .
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Illustrative
Mathematics
⎯⎯⎯⎯⎯⎯⎯
Here F is any point on AB (other than one of the endpoints) and
←→
←→
⎯⎯⎯⎯⎯⎯⎯⎯
G is the point on BC
so that FG parallel to line AC . Then we have the relationship
|FA|
|GC|
=
.
|AB|
|BC|
The converse of this result holds as well: if F is a point on
⎯⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯
AB and G is a point on BC
←→
←→
so that the above equality holds then then FG is parallel to AC .
The argument presented here for midpoints works also in the more general setting
mentioned in the above paragraph. Essential to this argument is the use of the AAA
criterion for triangle similarity, G-SRT.3. The task can be used for assessment provided
students are aware that they may use the AAA result in part (b) of this problem.
Edit this solution
Solution
a. We require two auxiliary points on line ℓ: a point D so that M is between
and a point E so that P is between M and E as in the picture below
←→
←→
P and D
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Illustrative
Mathematics
←→
←→
Lines ℓ and AC are parallel and AB is a transverse for these parallel lines. Therefore
angle CAB is congruent to angle DMA since these are alternate interior angles for a
transverse meeting parallel lines. Angle DMA is congruent to angle PMB by the
vertical angle theorem. So angle CAB is congruent to angle PMB.
The same argument will show that angles BPM and BCA are congruent: angle BCA is
congruent to angle EPC (alternate interior angles) and angle EPC is congruent to
angle BPM (vertical angles) establishing that angles BCA and BPM are congruent.
We have m(∠ABC) = m(∠MBP) since these are the same angles. Therefore the
triangles MBP and ABC share three congruent angles and they are similar triangles.
Alternatively, we could stop after producing two congruent angles and use the AA
criterion for similarity of triangles.
b. Since triangles MBP and ABC are similar it follows that
|PB|
|BM|
=
.
|CB|
|BA|
⎯⎯⎯⎯⎯⎯⎯
Since M is the midpoint of BA , we have |BM|
⎯⎯⎯⎯⎯⎯⎯⎯
is the midpoint of segment CB .
|BA|
=
1
and so |PB|
2
|CB|
=
1
which means that
2
P
G-SRT Joining two midpoints of sides of a triangle
Typeset May 4, 2016 at 20:45:06. Licensed by Illustrative Mathematics under a
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