OVERVIEW
Using Similarity and Proving Triangle Theorems
G.SRT.4
G.SRT.4
Prove theorems about triangles.
Theorems include: a line parallel to
one side of a triangle divides the
other two proportionally, and
conversely; the Pythagorean
Theorem proved using triangle
similarity.
(1) The student will prove (the side
splitting theorem) that a line parallel
to one side of a triangle divides the
other two proportionally.
(2) The student will prove (the angle
bisector theorem) that an angle
bisector of an angle of a triangle
divides the opposite side in two
segments that are proportional to the
other two sides of the triangle.
This objective introduces what is
sometimes known as the ‘side
splitting theorem’ or proportional
parts theorem. Look at the impact
that parallel lines have on dividing
the lines up proportionally.
Proportional values can be found
in other places in the triangle
when parallel lines are formed.
Focus on student outcome 1 more
than 2. The angle bisector has very
limited use outside of its own basic
problems. Emphasize the side
splitting theorem and its use in
multiple settings.
2 – The angle bisector theorem while
quite pretty to prove and quite easy
to use has very little application
beyond itself. The main focus here is
the ‘side splitting theorem.’
(3) The student will prove the
Pythagorean Theorem using similarity
and the geometric mean.
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1 – The most common error in this
unit is found in this objective –
Students often form a scale factor
between two proportional pieces and
then relate it to two proportional
sides. These two scale factors are not
equal and this causes many errors.
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NOTES
Using Similarity and Proving Triangle Theorems
G.SRT.4
CONCEPT 1 – Prove theorems about triangles – The Side Splitting Theorem - A line
parallel to one side of a triangle divides the other two proportionally.
This relationship is sometimes called the ‘Side Splitting” theorem.
Given:
A
DE || BC
E
AD AE

Prove:
DB EC
D
C
B
A
Given that DE || BC
mADE  mABC by Corresponding  
x
o
D
mAED  mACB by Corresponding  
E
x
C
o
Thus ADE  ABC by AA.
B
Similarity brings proportional sides….
AD AE

AB AC
A
Express the distance AB as the sum of its two pieces,
A
AB = AD + DB
o
o
D
*
E
*
B
(using segment addition)
and AC is the sum of its two pieces,
C
AC = AE + EC
(using segment addition)
So by substitution, it follows that:
AD
AE

AD  DB AE  EC
AD (AE + EC) = AE (AD + DB) by cross multiplication
AD(AE) + AD(EC) = AE(AD) + AE(DB) by distribution
Subtract AD(AE) from both sides leaving
AD(EC) = AE(DB)
AD AE


DB EC
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NOTES
Using Similarity and Proving Triangle Theorems
G.SRT.4
CONCEPT 2 – Prove theorems about triangles – The Angle Bisector Theorem – An
angle bisector of an angle of a triangle divides the opposite side in two segments that are
proportional to the other two sides of the triangle.
Given: ABC where BD is an angle bisector of B.
Prove:
B
**
AB AD

BC DC
A
Create an auxiliary parallel line to BD through Point A while also extending
C
D
E
B
side BC until the two meet at point E.
**
A
mE  mCBD because Corresponding Angles , and
mDBA  mEAB because Alternate Interior Angles 
E
*
D
C
B
*
* *
A
D
C
EB = AB because of the Isosceles Triangle Theorem
Thus using the side splitting theorem
EB AD
AB AD


and then using substitution
.
BC DC
BC DC
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NOTES
Using Similarity and Proving Triangle Theorems
G.SRT.4
CONCEPT 3 – Prove theorems about triangles. Prove the Pythagorean Theorem
using triangle similarity.
Given: A right triangle with an altitude (height) draw from
the right angle to the hypotenuse.
a
Prove: a 2  b 2  c 2
b
o
*
d
e
c
When comparing the left inner triangle to the entire
triangle, notice that both triangles have a common angle
and both have a right angle. Thus they are similar by AA.
Knowing that they are similar allows for the establishment
of the proportion between the sides….
left 
a d
 
whole c a
a
o
*
*
c
d
a 2  cd
When comparing the right inner triangle to the entire
triangle, notice that both triangles have a common angle
and both have a right angle. Thus they are similar by AA.
Knowing that they are similar allows for the establishment
of the proportion between the sides….
right  b e
 
whole c b
b
a
a
b
b
o
*
o
c
e
b 2  ce
Now the pretty part!!!
It is now known that
a 2  cd
and that
b 2  ce
The Pythagorean Theorem is beginning to appear…. Next, by adding both equations together.
Very close to the Pythagorean Theorem… Just factor out a c.
a 2  b2  c(d  e) from the diagram notice that length c = d + e and substitute that in….. WOW!!!
a 2  b2  c2
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ASSESSMENT
Using Similarity and Proving Triangle Theorems
G.SRT.4
1. In the given diagram, which of the following statement is NOT true:
A) ADE  ACB
A
B) ADE  AED
E
D
AD AE

C)
AC
AB
AD AE

D)
DC EB
B
C
2. In the given diagram, which of the following statement is NOT true:
A)
AB
AC

DE DF
B)
G
AB DE

EF
BC
D
A
E
B
AB DE

C)
BC EF
GB GE

D)
GA GD
F
C
3. Which of the following would not solve for the correct value of x:
8
5 x5
A) 
8
16
C)
4.
5 1
B) 
x 2
5 x 5

8
8
D)
x
5
x
5

16 8
x 10
x
10


is the same proportion as
.
3 15
x  3 25
T or F
5. Tim claims that because ADE  ACB, that the proportion
A
AD DE

is valid.
DC CB
E
D
Jennifer disagrees with Tim. Who is correct? Why?
B
C
If Jennifer is correct, rewrite the proportion so that it is true.
6. Complete the proportions.
a)
b)
AD DC
AC AG


AE
AD
d)
e)
DE AD

CB
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c)
A
EB

FG AG
AB
FG
AG
F
G
E
D
f)

8
B
AD AC

DF
C
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ASSESSMENT
Using Similarity and Proving Triangle Theorems
7. Find the values for the missing variables.
a)
b)
c)
d)
4
6
8
2.5
x
9
12
x
8.25
12
x
11
o
x = __________
o
3
8
x
x = __________
x = __________
2
x = __________
G.SRT.4
8. Determine the values for the variables, x and y.
4
6
x
24
y
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ASSESSMENT
Using Similarity and Proving Triangle Theorems
G.SRT.4
Answers:
1)
2)
3)
4)
B
B
A
T
5) Jennifer. The two similar triangles are ADE  ACE, which makes the correct proportion
AD DE

.
AC CB
The error that Tim made was that the measurements were all full lengths of sides except DC , it is only
AD( full )
DE ( full )

a portion of the side thus
is an incorrect proportion.
DC ( part ) CB( full )
6) a) EB
b) AF
c) AB
d) AC
e) EB
f) CG
7) a) x = 6.4
b) x = 11.25
c) x = 4
d) x = 16
4
x

8)
so x = 9.6. Use the Pythagorean Theorem to solve for y. 102  242  m2 therefore m = 26.
10 24
10 6
 results in y = 15.6.
Finally to solve for y using,
26 y
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