8-4 SIMILARITY IN RIGHT TRIANGLES (p. 439-444)
There is no need to do the Investigation.
B
B
B
C
D
C
A
C
A
D
A
D
C
When altitude CD is drawn to the hypotenuse of ABC in the first diagram on the left,
three similar triangles are formed. Let’s see why?
1. Why is BCA ~ CDA?
2. Why is BCA ~ BDC?
3. After answering the first two questions, why is CDA ~ BDC?
4. How do the three triangles relate to each other?
The first result from the investigation of the three right triangles is summarized in
Theorem 8-3.
Theorem 8-3
The altitude to the hypotenuse of a right triangle divides the triangle into two
triangles that are similar to the original triangle and to each other.
a c
 , b and c are called the means and a and d are called the
b d
extremes of the proportion.
In the proportion
a x
 , notice that the means are equal (they are both equal to x). This
x b
type of proportion is sometimes referred to as a mean proportion. Also, x is referred to as
the geometric mean of a and b (recall that a and b are the extremes).
By the cross-product property, x 2  ab. Why is x  ab ?
In the proportion
So, to find the geometric mean of two numbers, take the square root of the product of the
two numbers.
Example: Find the geometric mean of 2 and 27. Set up and solve a proportion. Leave
your answer in simplest radical form. How does the geometric mean differ from the
arithmetic mean (average) of two numbers? How do you find an arithmetic mean? Find
the arithmetic mean of 2 and 27. In this problem, which mean of 2 and 27 is larger?
Example: If 12 is the geometric mean of 3 and another number, find the other number.
Set up and solve a proportion.
Do 1 on p. 440.
Let’s now return to the three right triangles that we were introduced to in the beginning of
this section.
The second and third results from the investigation of the three right triangles are
summarized in corollaries to Theorem 8-3.
B
B
B
C
D
C
A
C
A
D
A
D
C
1. BCA ~ CDA
2. BCA ~ BDC
3. CDA ~ BDC
Using (3), why can we set up this proportion? What is the abbreviation for this
statement?
AD CD

Notice that both means are the same. This is a mean proportion. CD is the
CD BD
geometric mean between AD and BD (the two extremes). In other words, the altitude to
the hypotenuse of the given right triangle is the geometric mean between the segments of
the hypotenuse.
We have just proved the first corollary to Theorem 8-3.
Corollary 1 to Theorem 8-3
The length of the altitude to the hypotenuse of a right triangle is the geometric
mean of the lengths of the segments of the hypotenuse.
B
m ABC=90
m BDC=90
altitude
A
seg. D
of hyp.
C
segment of
hypotenuse
Here is a short way to remember this corollary:
hyp. seg. #1
alt.

alt.
hyp. seg. #2
B
B
B
C
D
C
A
C
A
D
A
D
C
1. BCA ~ CDA
2. BCA ~ BDC
3. CDA ~ BDC
Using (1), why can we set up this proportion? What is the abbreviation for this
statement?
AB AC

Notice that both means are the same. This is a mean proportion. AC is the
AC AD
geometric mean between AB and AD (the two extremes). In other words, the leg of the
given right triangle is the geometric mean between the hypotenuse of the given right
triangle and the segment of the hypotenuse adjacent to that leg.
B
B
B
C
D
C
A
C
A
D
A
D
C
1. BCA ~ CDA
2. BCA ~ BDC
3. CDA ~ BDC
Using (2), why can we set up this proportion?
AB CB

Notice that both means are the same. This is a mean proportion. CB is the
CB DB
geometric mean between AB and DB (the two extremes). In other words, the leg of the
given right triangle is the geometric mean between the hypotenuse of the given right
triangle and the segment of the hypotenuse adjacent to that leg.
We have just proved the second corollary to Theorem 8-3.
Corollary 2 to Theorem 8-3
Each leg of the original right triangle is the geometric mean of the length of the
adjacent hypotenuse segment and the length of the entire hypotenuse.
B
m ABC=90
m BDC=90
leg
A
D
adjace nt
hypotenuse
segment
C
m ABC=90
m BDC=90
B
leg
A
D
adjace nt
hypotenuse
segment
C
Here is a short way to remember this corollary:
adj. hyp. seg. leg

leg
hyp.
We will use these two corollaries to set up and solve proportions that involve geometric
means.
Example: Set up proportions to solve for x, y, and z. In the diagram,
m ABC  90 and m ADB  90. Leave answers in simplest radical form, if necessary.
B
y
A
z
x
4 D
21
C
After you found x in the above figure, what other theorem could you have used to find y
and z?
Example: Set up proportions to solve for x and y. In the diagram,
m ABC  90 and m ADB  90. Leave answers in simplest radical form, if necessary.
B
y
6
A
2 D
x
C
If time, do 2 on p. 441.
Example: At a golf course, Mary drove her ball 192 yd straight toward the cup from the
tee. Peter drove his ball straight 240 yd, but not toward the cup. Find x and y in the
following diagram, which represents their remaining distances from the cup. In the
diagram, m TPC  90 and m PMT  90. Also, after their drives, how far apart are their
golf balls?
C
y
x
z
P
M
192
240
T
Homework p. 442-444: 4,8,11-13,16,18,21,25,29,37,40,58,63,65,69
25. Sketch the points on graph paper. Since CD is the altitude to the hypotenuse, AB is
the hypotenuse. CD divides the hypotenuse into segments of lengths 4 and 9. So,
x 2  36 and x  6. Thus, C is at (4 + 6, 6) or (4 – 6, 6).
37. Solve
x
8

or 2x 2  64 and x  4 2.
8 2x
40. Use A 
1
bh
2
4 2  8 2  12 2
V
30 
10 3
20
60 
90 
B
10
A  50 3
1
(20) x
2
x  5 3 cm
50 3 
65. A  bh
A  6(10)  60
8h = 60
h = 7.5
A