9.1 Similar Right Triangles
Geometry
Mrs. Spitz
Spring 2006
Standard 3:

Students will understand geometric concepts
and applications.
Objectives/Assignment





Solve problems involving similar right triangles
formed by the altitude drawn to the hypotenuse of a
right triangle.
Use a geometric mean to solve problems such as
estimating a climbing distance.
Assignment: pp. 531-532 #1-34
Today’s assignment: Ch. 9 Definitions, Ch. 9
Postulates and Theorems, and Binder Check. I need
to see notes from 7.1, 7.2 (except 5th period), 8.1,
8.2, 8.3, 8.4, 8.5(except 2nd period), 8.6, and 8.7.
Algebra Review for extra credit is due by the end of
the week.
Proportions in right triangles

In Lesson 8.4, you learned
that two triangles are
similar if two of their
corresponding angles are
congruent. For example P
∆PQR ~ ∆STU. Recall
that the corresponding
side lengths of similar
triangles are in
proportion.
S
U
R
T
Q
Activity: Investigating similar right
triangles. Do in pairs or threes
1.
2.
3.
4.
Cut an index card along one of
its diagonals.
On one of the right triangles,
draw an altitude from the right
angle to the hypotenuse. Cut
along the altitude to form two
right triangles.
You should now have three
right triangles. Compare the
triangles. What special
property do they share?
Explain.
Tape your group’s triangles to a
piece of paper and place in
labwork.
What did you discover?

In the activity, you may have discovered the
next theorem. A plan for proving the theorem
appears on page 528. You are asked to prove
the theorem in Exercise 34 on page 533.
Theorem 9.1

If the altitude is drawn
to the hypotenuse of a
right triangle, then the
two triangles formed
are similar to the
original triangle and to
A
each other.
C
D
∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD
B
A plan for proving thm. 9.1 is shown
below:



Given: ∆ABC is a right triangle; altitude CD is drawn to
hypotenuse AB.
Prove: ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD
Plan for proof: First prove that ∆CBD ~ ∆ABC. Each
triangle has a right triangle and each includes B. The
triangles are similar by the AA Similarity Postulate. You can
use similar reasoning to show that ∆ACD ~ ∆ABC. To show
that ∆CBD ~ ∆ACD, begin by showing that ACD  B
because they are both complementary to DCB. Then you
can use the AA Similarity Postulate.
C
A
D
B
Ex. 1: Finding the Height of a Roof



Roof Height. A roof has a
cross section that is a right
angle. The diagram shows
the approximate
dimensions of this cross
section.
A. Identify the similar
triangles.
B. Find the height h of the
roof.
Solution:

Y
You may find it helpful to
sketch the three similar
3.1 m
h
triangles so that the
corresponding angles and X
W
sides have the same
orientation. Mark the
congruent angles. Notice
that some sides appear in
5.5 m
more than one triangle.
For instance XY is the
hypotenuse in ∆XYW and
the shorter leg in ∆XZY.
Y
h
∆XYW ~ ∆YZW ~ ∆XZY.
Z
6.3 m
Z
X
W
3.1 m
5.5 m
Y
Solution for b.

Use the fact that ∆XYW ~ ∆XZY to write a
proportion.
YW
ZY
=
XY
XZ
Corresponding side lengths are in
proportion.
h
5.5
=
3.1
6.3
Substitute values.
6.3h = 5.5(3.1)
h ≈ 2.7
Cross Product property
Solve for unknown h.
The height of the roof is about 2.7 meters.
Using a geometric mean to solve
problems

In right ∆ABC,
altitude CD is drawn
to the hypotenuse,
forming two smaller
right triangles that
are similar to ∆ABC
From Theorem 9.1,
you know that ∆CBD
~ ∆ACD ~ ∆ABC.
C
A
D
B
B
C
D
C
D
A
A
B
C
Write this down!
C
A
D
B
B
C
D
C
A
D
B
Notice that CD is the longer leg
of ∆CBD and the shorter leg
of ∆ACD. When you write a
proportion comparing the
legs lengths of ∆CBD and
∆ACD, you can see that CD
is the geometric mean of BD
and AD.
Longer leg of ∆CBD.
Shorter leg of ∆CBD.
BD
CD
A
C
Shorter leg of ∆ACD
=
CD
AD
Longer leg of ∆ACD.
Copy this down!
C
A
D
B
B
C
D
C
A
Sides CB and AC also appear in
more than one triangle. Their
side lengths are also geometric
means, as shown by the
proportions below:
D
B
BD
CD
Hypotenuse of ∆CBD
A
Shorter leg of ∆ABC.
Hypotenuse of ∆ABC.
C
=
CD
AD
Shorter leg of ∆CBD.
Copy this down!
C
A
D
B
B
C
D
C
A
Sides CB and AC also appear in
more than one triangle. Their
side lengths are also geometric
means, as shown by the
proportions below:
D
B
BD
CD
Hypotenuse of ∆ACD
A
Longer leg of ∆ABC.
Hypotenuse of ∆ABC.
C
=
CD
AD
Longer leg of ∆ACD.
Geometric Mean Theorems


Theorem 9.2: In a right triangle, the
altitude from the right angle to the
hypotenuse divides the hypotenuse into
two segments. The length of the
altitude is the geometric mean of the
lengths of the two segments
Theorem 9.3: In a right triangle, the
altitude from the right angle to the
hypotenuse divides the hypotenuse into
two segments. The length of each leg
of the right triangle is the geometric
mean of the lengths of the hypotenuse
and the segment of the hypotenuse that
is adjacent to the leg.
C
A
D
BD
=
CD
CD
AD
AB
CB
=
CB
DB
AB
AC
=
AC
AD
B
What does that mean?
2
x
6
6
=
x
18 = x2
√18 = x
5
y
3
x
5+2
3
y
7
y
=
y
=
2
y
2
√9 ∙ √2 = x
14 = y2
3 √2 = x
√14 = y
Ex. 3: Using Indirect Measurement.

MONORAIL TRACK. To
estimate the height of a
monorail track, your friend
holds a cardboard square at
eye level. Your friend lines up
the top edge of the square with
the track and the bottom edge
with the ground. You measure
the distance from the ground
to your friend’s eye and the
distance from your friend to
the track.
In the diagram, XY = h – 5.75 is the difference between the track
height h and your friend’s eye level. Use Theorem 9.2 to write a
proportion involving XY. Then you can solve for h.
Upcoming:



March 17 is the last day of the Third Quarter.
Grades are due soon after.
There are quizzes after 9.3 and 9.5
Test should be Wednesday or Thursday before
you go on Spring Break – don’t miss it. If
you are leaving early somewhere, take it
before you leave.