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In this presentation, we will understand the different rules of congruence and similarity of triangles with some more basic rules that will be applied to the same.
1.
Chapter 14
Deductive Geometry-
Geometry Using Triangle
Congruence and Similarity
2.
Review: Parallel Lines and Angles
Two different given lines L1 and L2 on a plane are said to be parallel if
they will never intersect each other no matter how far they are
Two angles are called vertical angles if they are opposite to each other
and are formed by a pair of intersecting lines.
A B
Any pair of vertical angles are always congruent.
3.
Parallel Lines and Angles
Given two line L1 and L2 (not necessarily parallel) on the plane, a third line T
is called a transversal of L1 and L2 if it intersects these two lines.
L1
L2
T
4.
Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T
be a transversal.
a) a and form a pair of corresponding angles.
b) c and form a pair of corresponding angles etc.
L1
a
c
L2
T
5.
Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T
be a transversal.
c) c and form a pair of alternate interior angles.
d) d and form a pair of alternate interior angles.
L1
c
d L2
T
6.
Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T
be a transversal.
e) a and form a pair of alternate exterior angles.
f) b and form a pair of alternate exterior angles.
L1
a
L2
T
7.
Let L1 and L2 be two lines on the plane, and T be a transversal.
If L1 and L2 are parallel, then
a) any pair of corresponding angles are congruent,
b) any pair of alternate interior angles are congruent,
c) any pair of alternate exterior angles are congruent.
L1
L2
T
8.
Let L1 and L2 be two lines on the plane, and T be a transversal.
a) if there is a pair of congruent corresponding angles, then L1 and L2
are parallel.
b) if there is a pair of congruent alternate interior angles, then L1 and L2
are parallel.
c) if there is a pair of congruent alternate exterior angles, then L1 and L2
are parallel. L
1
L2
T
9.
Section 14.1 Congruence of Triangles
Given two triangles ΔABC and ΔXYZ.
If AB is congruent to XY, A is congruent to X ,
BC is congruent to YZ, B is congruent to Y ,
CA is congruent to ZX, C is congruent to Z
then we say that ΔABC is congruent to ΔXYZ, and we write
ABC XYZ
Y
B
X Z
C
10.
Congruence of Triangles
Side-Angle-Side Principle
Given two triangles ΔABC and ΔXYZ.
If AB is congruent to XY
B is congruent to Y
BC is congruent to YZ Z
then ΔABC is congruent to ΔXYZ
B
Y
C
X
11.
Congruence of Triangles
Angle-Side-Angle Principle
Given two triangles ΔABC and ΔXYZ.
If A is congruent to X
AC is congruent to XZ
C is congruent to Z Z
then ΔABC is congruent to ΔXYZ
((
B
Y
((
C (
)
X
12.
Congruence of Triangles
Side-Side-Side Principle
Given two triangles ΔABC and ΔXYZ.
If AB is congruent to XY
BC is congruent to YZ
CA is congruent to ZX Z
then ΔABC is congruent to ΔXYZ
B
Y
C
X
13.
Congruence of Triangles
Hypotenuse-Leg (HL) Principle
Given two right triangles ΔABC and ΔXYZ.
If hypotenuse of ΔABC is congruent to hypotenuse of ΔXYZ,
one leg of ΔABC is congruent to one leg of ΔXYZ,
then ΔABC is congruent to ΔXYZ
Z
A
8
13 Y
13
C X
B 8
14.
Theorem
If ΔABC is congruent to ΔXYZ , then
AB is congruent to XY
BC is congruent to YZ
CA is congruent to ZX
and
A is congruent to X
B is congruent to Y
C is congruent to Z
In short, corresponding parts of congruent triangles
are congruent. This is sometimes referred to as the
“CPCTC” property.
15.
Section 14.2 Similarity of Triangles
Given ΔABC and ΔXYZ.
If A is congruent to X
B is congruent to Y
C is congruent to Z
and AB : XY = BC : YZ = CA : ZX
then we say that ΔABC is similar to ΔXYZ, and the notation is
ΔABC ~ ΔXYZ
Y
B
X Z
A C
16.
Similarity of Triangles
SSS similarity principle
Given ΔABC and ΔXYZ.
If AB : XY = BC : YZ = CA : ZX
then ΔABC is similar to ΔXYZ.
Y
Z
B
A C
X
17.
Similarity of Triangles
AAA similarity principle
Given ΔABC and ΔXYZ.
If A is congruent to X
B is congruent to Y
C is congruent to Z
then ΔABC is similar to ΔXYZ
X
B
A C Z
Y
18.
Similarity of Triangles
AA similarity principle
Given ΔABC and ΔXYZ.
If A is congruent to X
B is congruent to Y
then ΔABC is similar to ΔXYZ
(because the angle sum of a triangle is always 180o)
X
B
A C Z
Y
19.
Similarity of Triangles
SAS similarity principle
Given ΔABC and ΔXYZ.
If AB : XY = BC : YZ and
B is congruent to Y
then ΔABC is similar to ΔXYZ
Y
B
A C
X Z
20.
Indirect Measurements
If the shadow of a tree is 37.5 m long, and the shadow of
a 1.5m student is 2.5 m long. How tall is the tree?
(assuming that they are all on level ground.)
21.
Indirect Measurements
What is the distance between
the two points A and B on the A
rim of the pond?
47m
B
58m
40
o
C
22.
Indirect Measurements
How far is the boat from point B
A on the shore?
How do we measure angles?
72º
68º 36m
C land A
23.
Artillery Rangefinder
25.
A large (3.5m) optical rangefinder mounted on the flying bridge of
the USS Stewart (a Destroyer Escort)
26.
A Transit
is a surveying instrument to
measure horizontal angles.
Glass reticle on both
models has stadia lines
for measuring distance.
Stadia ratio 1:100
28.
Measure the height of Devils Tower in Wyoming
29.
In the field, measure the angles and the distance.
Devils Tower
On paper, create a similar triangle using AA similarity principle.
Measure the height h (in cm); and when we scale it back, we can get the
height of the real tower.
30.
The first 'proper' ascent was in 1937 when some of America's best climbers took on
the project. The Weissner Route was the result, a 5.7 (decent VS) classic on which
he placed a single per runner on the crux pitch.
31.
A bizarre incident took place
only a couple of years later, in
1941 when local air ace
Charles George Hopkins
decided to parachute onto the
top of the tower to advertise
his aerial show. He came
prepared with a length of rope,
a block and tackle as well as a
sharpened axle from a Model
T Ford to act as an anchor for
his planned escape.
His parachute descent went OK but on arrival he found that his rope had
disappeared over the edge and he was well and truly stuck! There was only one
solution, the first ascensionists were called upon to drive halfway across America
to repeat their great feat and bring down the hapless Hopkins who had spent a cold
and lonely week on his island in the sky.
32.
Example 1. Use similar triangles to find the value of x.
x
7
33.
Example 2. Use similar triangles to find the values of x and y.
x
y
34.
Example 2. Use similar triangles to find the values of x and y.
R
4.
5
x
S
x 8
Q
y
35.
Example 2. Use similar triangles to find the values of x and y.
R R
5 4.
4. 5
x x
S S
x 88
Q Q
y
36.
Example 2. Use similar triangles to find the values of x and y.
R
R 5
4.
x x
S
4.5
8
x Q
S 8 Q
y
37.
Example 3. Use similar triangles to find the values of x and y.
y
x
38.
Example 3. Use similar triangles to find the values of x and y.
C
4
x D
9
y A
x
39.
Example 3. Use similar triangles to find the values of x and y.
CC
4 4
D x x D
9 9
y A A
x
40.
Example 3. Use similar triangles to find the values of x and y.
A
C
4
D x
9 x
9 D
y C
4
A
x
41.
Example 4. Use similar triangles to find the value of x.
C
x
A B
12 D 3
42.
C C
x x
C
D D 3 B
A 12
x
A B
12 D 3
43.
C C
x x
C
D D 3 B
A 12
x
A B
12 D 3
44.
D D
C
12 x
x 12
x
AC CA
C
D 3 B
x
A B
12 D 3
45.
D
C
12 x
x
A C
C
D 3 B
x
A B
12 D 3
46.
D
D
D
x
x
3
3
12 x
B
B
C
C
A C
C
x
A B
12 D 3
47.
Section 14.3
Basic Euclidean Constructions
(with Straight Edge and Compass)
A straight edge is only used to draw line segments, and it should not have
any marking on it. If a ruler is used, then all the markings should be
48.
Reasons for not using a
Protractor or Ruler
1. We need to prepare for situations where some equipments
are not available.
2. Small protractors are not accurate enough for large scale
projects.
3. Many geometric constructions cannot be done by protractors
but can be done by a compass and straight edge, such as
finding the center of a circle.
49.
What can the equipments do?
1. A straight edge is infinitely long and is only used to draw
line segments connecting given points or extend an
existing line segment.
2. A compass can be used to draw circles and circular arcs
with radii that have already been constructed.
50.
Basic Rules
Any construction should consist of only a finite number of steps.
Only one step can be carried out at a time.
Each construction must be exact. Approximation is not counted as
a solution. In particular, a construction should not require drawing
a line tangent to an existing circular arc because that will not
provide accurate results.
A point must either be given in advance, or be constructed by the
intersection of two line segments, or two arcs, or an arc and a line
51.
Basic Rules
All compass and straightedge constructions consist of repeated application of
five basic constructions using the points, lines and circles that have already
been constructed or given. These are:
1. Creating the line through two existing points
2. Creating the circle through one point with centre at another point.
3. Creating the point which is the intersection of two existing, non-
parallel lines.
4. Creating the one or two points in the intersection of a line and a
circle.
5. Creating the one or two points in the intersection of two circles.
52.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step I
Adjust the compass until
the gap between the
pencil and needle is
more than half the
length AB.
Move the needle tip of
the compass to point A
A B
53.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step II
Draw an approx. half circle as shown (click)
A B
54.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step II
Draw an approx. half circle as shown (click)
A B
55.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step III
Move the needle tip to point B. (click)
A B
56.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step III
Move the needle tip to point B. (click)
A B
57.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step III
Move the needle tip to point B. (click)
A B
58.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step III
Move the needle tip to point B. (click)
A B
59.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step IV
Draw an approx. half circle as shown (click)
A B
60.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step IV
Draw an approx. half circle as shown
A B
61.
The following animation shows how to construct the perpendicular
bisector (of a line segment) using just a straight edge and compass.
Step V
Use a ruler and a pencil to draw a line connecting the two
intersections of the circular arcs.
This will be the perpendicular bisector of AB.
A B
62.
Standard examples
1. Construction of an equilateral triangle, hence the construction of a
60o angles
2. Construction of a right angle.
3. Construction of a square.
4. Bisect a given angle.
5. Construction of a line segment parallel to a given line segment.
6. Given a unit segment and any positive integer n, construction of a
line segment whose length is n units.
7. Divide any given line segment into any given (whole) number of
congruent parts.
8. Locating the exact centre of a given circle.
9. Given a unit segment and any positive real number x, construction of
a segment whose length is square root of x times the unit segment.
10. Given a unit segment and any positive real number x, construction
of a segment whose length is 1/x times the unit segment.
63.
Construction of Regular Polygons
(by compass and straight edge)
1. Is it possible to construct an equilateral
triangle? Yes
2. Is it possible to construct a square? Yes
3. Is it possible to construct a regular pentagon? Yes
4. Is it possible to construct a regular hexagon? Yes
5. Is it possible to construct a regular heptagon? No
6. Is it possible to construct a regular octagon? Yes
64.
Important Theorem
A regular n-sided polygon is constructible if and only if
one of the following is true.
(1) n is a power of two (but bigger than 2), or
(2) n is an odd prime from the list 3, 5, 17, … (where the
k
general form is 22 + 1 for some whole number k.)
(3) n is of the form
2mp1p2…pr
where m is a whole number and pi’s are distinct odd
primes of the above form.
Examples:
A regular polygon with 30 sides is constructible, but one with 35 sides
will not be constructible. A regular polygon of 36 sides is also not
constructible.
66.
What kind of regular polygons are constructible?
1. Any equilateral triangle is constructible.
2. Any square is constructible.
3. Any regular pentagon is constructible. (see lab 20)
4. Any regular hexagon is constructible.
5. Is a regular heptagon constructible? Why?
No, because 7 is a prime not on the list of the previous theorem.
6. Is a regular octagon constructible? Why?
Yes, because 8 is a power of 2, and the previous theorem says
that it can be done.
67.
What kind of regular polygons are constructible?
7. Is a regular nonagon constructible? Why?
No, b/c 9 = 3×3, the 3 is repeating.
8. Is a regular decagon constructible? Why?
Yes, b/c 10 = 2×5, and 5 is on the list.
9. Is a regular 11-gon constructible? Why?
No, b/c 11 is a prime but not on the list.
10. Is a regular 12-gon constructible? Why?
Yes, b/c 12 = 2×2×3 and 3 is on the list.
11. Is a regular 13-gon constructible? Why?
No, b/c 13 is a prime but not on the list.
12. Is a regular 14-gon constructible? Why?
No, b/c 14 = 2×7 but 7 is an odd prime not on the list.
68.
Constructible Angles
We know that angles with measures of 90°, 60°, 45°, 30° are all constructible
by straight edge and compass (see Lab 20).
1. Is it possible to construct a 75° angle (with straight edge and compass)?
Yes, b/c 75° = 45° + 30°
2. Is it possible to construct a 15° angle (with straight edge and compass)?
Yes, b/c 15° = 45° – 30°, and 15° = 30° ÷ 2
69.
3. Is it possible to construct a 72° angle (with straight edge and compass)?
Yes, b/c a regular pentagon is constructible and its central angle has
measure equal to 72°.
g le
an
l
ra
nt
ce
center
Central angle = 360° ÷ 5 = 72°
4. Is it possible to construct a 3° angle (with straight edge and compass)?
Yes, b/c 75° – 72° =3°
70.
5. Is it possible to construct a 40° angle (with straight edge and compass)?
No, b/c the central angle θ of a regular
nonagon is exactly 360° ÷ 9 = 40°, and
from a previous result, we know that a
θ regular nonagon is not constructible.
6. Is it possible to construct a 1° angle (with straight edge and compass)?
No, b/c otherwise we can construct a 40° angle by repeated
addition of angles.
71.
I. The smallest constructible angle with positive integer degree
measure is a 3° angle.
II. If n is a positive integer, and angle A has measure n°, then angle
A is constructible if and only if n is a multiple of 3.
Conclusion II above does not apply to angles with non-integer degree
measures. For instance, a 7.5° angle is constructible, but 7.5 is not a
multiple of 3.
Example: Can we construct an 14° angle with just straight edge and
compass?
72.
What cannot be done?
1. Trisect an angle whose measure is not 3 times of an angle
which is already constructible by another method, for
instance, 90o can be trisected because a 30o angle is
constructible by bisecting a 60o angle. However a 30°
angle cannot be trisected.
2. Square a circle: i.e. construct a square with the same area
as a given circle.
(consequently, circling a square is also impossible.)
3. Construct a line segment whose length is
3
2times a given unit line segment.
73.
Enrichment
Rectangle to Square Dissection Problem
Challenge: Can we dissect an arbitrary rectangle into a finite number of pieces and
then rearrange those pieces into a square with exactly the same area?
The answers is yes, and it requires only 3 pieces if its length is no more than twice of
its width. The following animation shows how this is done.
A Step 1
Draw a square with exactly the same
area as shown. (click)
Step 2
Construct the line segment AB.
(click)
Click to see step 3
B
74.
Rectangle to Square Dissection Problem
Challenge: Can we dissect an arbitrary rectangle into a finite number of pieces and
then rearrange those pieces into a square with exactly the same area?
The answers is yes, and it requires only 3 pieces if its length is no more than 4 times
of its width. The following animation shows how this is done.
Step 1
Draw a square with exactly the same
area as shown. (click)
Step 3
Slide the triangles up. (click)
Step 2
Construct the line segment AB.
(click)
75.
If the length is more than 4 times of the width, we first cut the rectangle into
two congruent shorter pieces and then rearrange them into a rectangle with
shorter length. (click to see animation)
76.
Enrichment
Tarski’s circle-squaring problem
posed by Alfred Tarski in 1925, is to take a circle (including its interior) in
the plane, cut it into finitely many pieces, and reassemble the pieces so as to
get a square of exact area.
(click to see what will happen if we put the circle on top of the square.)
77.
A partial solution was given by Miklos Laczkovich in 1990; whose
decomposition consists of about 1050 different pieces, but the pieces do not
have nice boundaries (i.e. not Jordan curves) and hence cannot be cut out
with scissors.
It might be possible that some day, some one will give a better solution to
this problem.
78.
Section 14.5
Geometric Problem Solving using
Triangle Congruence and Similarity
In this section, we apply triangle congruence and similarity to
prove properties of geometric shape. Many of these properties
have been observed informally in Chapter 12.
79.
Loosely speaking, a mathematical proof is a valid logical argument for
a mathematical statement.
A two column proof is an argument presented in a table with 2 columns,
where statements and reasons are numbered to show their order in the
argument.
80.
A proof can also be written in paragraph form, which is call a paragraph
Paragraph proofs may be harder for beginners to read, but they are more
versatile and expressive, hence they are widely used by mathematicians.
Theorem 6.2
81.
Example 14.5
Prove that the diagonals in a kite are perpendicular to each other.
Statements Reasons
A
1 2
B D
E
C
Note: The reflexive property of congruence says that any object is congruent to itself.
82.
A kite can never be a trapezoid.
Statements Reasons
1. Given conditions of a kite.
2. Assuming that ABCD can be a
trapezoid.
3. Alternate interior angles of
parallel lines are congruent.
4. SSS congruence principle.
5. Corresponding parts of
congruent triangles are
congruent.
6. Combining (3) and (5).
7. Transitive property of equality
8. Congruence of alternate interior
angles implies parallel lines.
9. A trapezoid has exactly one pair
of parallel opposite sides.
83.
Statements Reasons
1. Given
2. The same angle.
3. Given
4. SAS similarity principle
5. Corresponding angles of
similar triangles are congruent.
6. Converse of corresponding
angle theorem for parallel lines.
7. Corresponding sides of similar
triangles are in same ratio.
84.
We call PQRS the midquad of ABCD because it is a
quadrilateral formed by joining the midpoints of the
sides of ABCD.
85.
Statements Reasons
1. For any two points, we can
connect a line segment.
2. Midsegment theorem for ABC.
3. Midsegment theorem for ADC.
4. Transitive property for parallel
lines.
5. Similar reasons as (1) to (4).
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