Download congruent triangles 6.1.1 – 6.1.4

CONGRUENT TRIANGLES
6.1.1 – 6.1.4
Two triangles are congruent if there is a sequence of rigid transformations that carry one
onto the other. Two triangles are also congruent if they are similar figures with a ratio of
similarity of 1, that is 11 . One way to prove triangles congruent is to prove they are
similar first, and then prove that the ratio of similarity is 1. In these lessons, students find
short cuts that enable them to prove triangles congruent in fewer steps, by developing
five triangle congruence conjectures. They are SSS ≅, ASA ≅, AAS ≅, SAS ≅, and HL ≅,
illustrated below.
ASA ≅
SSS ≅
SAS ≅
AAS ≅
HL ≅
Note: “S” stands for “side” and “A” stands for “angle.” HL ≅ is only used with right
triangles. The “H” stands for “hypotenuse” and the “L” stands for leg. The pattern
appears to be “SSA” but this arrangement is NOT one of our conjectures, since it is only
true for right triangles.
See the Math Notes boxes in Lessons 6.1.1 and 6.1.4.
76
Core Connections Geometry
Example 1
Use your triangle congruence conjectures to decide whether or not each pair of triangles must be
congruent. Base each decision on the markings, not on appearances. Justify each answer.
a.
b.
c.
d.
e.
f.
In part (a), the triangles are congruent by the SAS ≅ conjecture. The triangles are also congruent
in part (b), this time by the SSS ≅ conjecture. In part (c), the triangles are congruent by the
AAS ≅ conjecture. Part (d) shows a pair of triangles that are not necessarily congruent. The first
triangle displays an ASA arrangement, while the second triangle displays an AAS arrangement.
The triangles could still be congruent, but based on the markings, we cannot conclude that they
definitely are congruent. The triangles in part (e) are right triangles and the markings fit the
HL ≅ conjecture. Lastly, in part (f), the triangles are congruent by the ASA ≅ conjecture.
Parent Guide with Extra Practice
77
Example 2
Using the information given in the diagrams below, decide if any triangles are congruent, similar
but not congruent, or not similar. If you claim the triangles are congruent or similar, create a
flow chart justifying your answer.
B
a.
b.
Z
X
V
D
Y
W
A
C
In part (a), ΔABD ≅ ΔCBD by the SAS ≅ conjecture. Note: If you only see “SA,” observe that
BD is congruent to itself. The Reflexive Property justifies stating that something is equal or
congruent to itself.
AD = CD
given
given
BD = BD
Reflexive prop.
ΔABD ≅ ΔCBD
by
SAS ≅
In part (b), ΔWXV ~ ΔZYV by the AA ~ conjecture. The triangles are not necessarily congruent;
they could be congruent, but since we only have information about angles, we cannot conclude
anything else.
Vertical angles =
ΔWXV ~ ΔZYV
AA ~
Lines , alt. int. angles =
There is more than one way to justify the answer to part (b). There is another pair of alternate
interior angles (∠WXV and ∠ZYV) that are equal that we could have used rather than the vertical
angles, or we could have used them along with the vertical angles.
78
Core Connections Geometry
Problems
Briefly explain if each of the following pairs of triangles are congruent or not. If so, state the
triangle congruence conjecture that supports your conclusion.
A F
1.
E
B
C
D
I
M
L
W
V
5.
P
3.
K
H
Q
4.
G J
2.
E
A
6.
D
Y
R
7.
H
U
8.
I
G
T
T
S
C
B
9. P
Q
X
L
N
K
J
O
N
O
M
R
S
10. T
U
B
11. A
X
14.
K
39º
P 112º
H
I
4
V
R
13.
3
5
D
C
W
G
12. E
12
5
13
F
39º
L
A
W
12
8
112º
J
15.
B
C4
D
4 F
12
Z
X
8
E
Q
Y
Use your triangle congruence conjectures to decide whether or not each pair of triangles must be
congruent. Base your decision on the markings, not on appearances. Justify your answer.
16.
A
B
E
C
17.
P
D
Q
Parent Guide with Extra Practice
S
R
79
18.
V
E
19.
Y
S
X
T
W
20.
A
Z
K
L
B
21.
N
O
E
R
A
Using the information given in each diagram below, decide if any triangles are congruent, similar
but not congruent, or not similar. If you claim the triangles are congruent or similar, create a
flowchart justifying your answer.
22.
D
A
23.
U
L
V
S
N
C
I
H
T
24.
25.
R
A
P
J
K
E
80
I
S
Core Connections Geometry
In each diagram below, are any triangles congruent? If so, prove it. (Note: Justify some using
flowcharts and some by writing two-column proofs.)
B
26.
27.
28.
D
B
B
C
A
A
A
E
C
D
29.
30.
D
C
D
D
C
E
31.
C
B
A
F
A
B
B
D
C
A
Complete a proof for each problem below in the style of your choice.
32.
Given: TR and MN bisect each other.
Prove: ΔNTP ≅ ΔMRP
33.
Given: CD bisects ∠ACB; ∠1 ≅ ∠2.
Prove: ΔCDA ≅ ΔCDB
A
R
N
P
M
T
1
C
2
D
B
34.
Given: AB CD , ∠B ≅ ∠D, AB ≅ CD
Prove: ΔABF ≅ ΔCDE
35.
Given: PG ≅ SG , TP ≅ TS
Prove: ΔTPG ≅ ΔTSG
C
P
E
B
S
A
36.
Given: OE ⊥ MP , OE bisects ∠MOP
Prove: ΔMOE ≅ ΔPOE
M
O
G
T
D
F
37.
Given: AD BC , DC BA
Prove: ΔADB ≅ ΔCBD
D
C
E
P
Parent Guide with Extra Practice
A
B
81
38.
Given: AC bisects DE , ∠A ≅ ∠C
Prove: ΔADB ≅ ΔCEB
39.
Given: PQ ⊥ RS , ∠R ≅ ∠S
Prove: ΔPQR ≅ ΔPQS
R
D
C
P
B
A
Q
E
S
40.
Given: ∠S ≅ ∠R, PQ bisects ∠SQR
Prove: ΔSPQ ≅ ΔRPQ
41.
S
P
Given: TU ≅ GY , KY HU , KT ⊥ TG ,
HG
⊥ TG . Prove: ∠K ≅ ∠H
K
U
T
Q
Y
H
R
42.
Given: MQ WL , MQ ≅ WL
Prove: ML WQ
Q
W
L
M
Consider the diagram at right.
43.
Is ΔBCD ≅ ΔEDC? Prove it!
44.
Is AB ≅ DC ? Prove it!
45.
Is AB ≅ ED ? Prove it!
82
G
B
A
D
C
E
Core Connections Geometry
Answers
1.
ΔABC ≅ ΔDEF by ASA.
2.
ΔGIH ≅ ΔLJK by SAS.
3.
ΔPNM ≅ ΔPNO by SSS.
4.
QS ≅ QS , so ΔQRS ≅ ΔQTS by HL.
5.
The triangles are not necessarily congruent.
6.
ΔABC ≅ ΔDFE by ASA or AAS.
7.
GI ≅ GI , so ΔGHI ≅ ΔIJG by SSS.
8.
Alternate interior angles = used twice, so ΔKLN ≅ ΔNMK by ASA.
9.
Vertical angles ≅ at 0, so ΔPOQ ≅ ΔROS by SAS.
10. Vertical angles and/or alternate interior angles =, so ΔTUX ≅ ΔVWX by ASA.
11.
No, the length of each hypotenuse is different.
12.
Pythagorean Theorem, so ΔEGH ≅ ΔIHG by SSS.
13.
Sum of angles of triangle = 180º, but since the equal angles do not correspond, the triangles
are not congruent.
14.
AF + FC = FC + CD, so ΔABC ≅ ΔDEF by SSS.
15.
XZ ≅ XZ , so ΔWXZ ≅ ΔYXZ by AAS.
16.
ΔABC ≅ ΔEDC by AAS ≅
17.
ΔPQS ≅ ΔPRS by AAS ≅, with PS ≅ PS by the Reflexive Property.
18.
ΔVXW ≅ ΔZXY by ASA ≅, with ∠VXW ≅ ∠ZXY because vertical angles are ≅.
19.
ΔTEA ≅ ΔSAE by SSS ≅, with EA ≅ EA by the Reflexive Property.
20.
ΔKLB ≅ ΔEBL by HL ≅, with BL ≅ BL by the Reflexive Property.
21.
Not necessarily congruent.
22.
ΔDAV ~ ΔISV by SAS ~
Equal
markings
Vert. angles =
ΔDAV ~ ΔISV
SAS~
23.
ΔLUN and ΔHTC are not necessarily similar based on the markings.
Parent Guide with Extra Practice
83
24.
ΔSAP ~ ΔSJE by AA~.
AP || JE
given
Alt. int.
angles =
Alt. int.
angles =
ΔSAP ~ ΔSJE
AA~
25.
ΔKRS ≅ ΔISR by HL ≅
given
ΔKRS & ΔISR are right
triangles.
given
KR = IS
RS = RS
ΔKRS
≅ ΔISR
Refl. Prop.
HL≅
26. Yes
∠BAD ≅ ∠BCD
Given
∠BDC ≅ ∠BDA
27. Yes
right ∠'s are ≅
BD ≅ BD
∠BCA ≅ ∠BCD
∠B ≅ ∠ E
Given
vertical ∠s are ≅
BC ≅ CE
Reflexive
Given
Δ ABD ≅ Δ CBD
Δ ABC ≅ Δ DEC
AAS
ASA
28. Yes
AC ≅ CD
Given
∠ BCD ≅ ∠ BCA
right ∠'s are =
BC ≅ BC
29. Yes
AD ≅ BC
Given
Reflexive
BA ≅ CD
CA ≅ CA
Given
Reflexive
Δ ABC ≅ Δ DBC
Δ ABC ≅ Δ CDA
SAS
SSS
30.
Not necessarily.
Counterexample:
31.
Yes
BC ≅ EF
Given
AC ≅ FD
Given
Δ ABC ≅ Δ DEF
HL
32.
NP ≅ MP and TP ≅ RP by definition of bisector. ∠NPT ≅ ∠MPR because vertical angles
are equal. So, ΔNTP ≅ ΔMRP by SAS.
33.
∠ACD ≅ ∠BCD by definition of angle bisector. CD ≅ CD by reflexive so ΔCDA ≅ ΔCDB
by ASA.
34.
∠A ≅ ∠C since alternate interior angles of parallel lines congruent so ΔABF ≅ ΔCDE by
ASA.
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Core Connections Geometry
35.
TG ≅ TG by reflexive so ΔTPG ≅ ΔTSG by SSS.
36.
∠MEO ≅ ∠PEO because perpendicular lines form ≅ right angles ∠MOE ≅ ∠POE by angle
bisector and OE ≅ OE by reflexive. So, ΔMOE ≅ ΔPOE by ASA.
37.
∠CDB ≅ ∠ABD and ∠ADB ≅ ∠CBD since parallel lines give congruent alternate interior
angles. DB ≅ DB by reflexive so ΔADB ≅ ΔCBD by ASA.
38.
DB ≅ EB by definition of bisector. ∠DBA ≅ ∠EBC since vertical angles are congruent.
So ΔADB ≅ ΔCEB by AAS.
39.
∠RQP ≅ ∠SQP since perpendicular lines form congruent right angles. PQ ≅ PQ by
reflexive so ΔPQR ≅ ΔPQS by AAS.
40.
∠SQP ≅ ∠RQP by angle bisector and PQ ≅ PQ by reflexive, so ΔSPQ ≅ ΔRPQ by AAS.
41.
∠KYT ≅ ∠HUG because parallel lines form congruent alternate exterior angles.
TY + YU = YU + GU so TY ≅ GU by subtraction. ∠T ≅ ∠G since perpendicular lines form
congruent right angles. So ΔKTY ≅ ΔHGU by ASA. Therefore, ∠K ≅ ∠H since ≅ triangles
have congruent parts.
42.
∠MQL ≅ ∠WLQ since parallel lines form congruent alternate interior angles. QL ≅ QL by
reflexive so ΔMQL ≅ ΔWLQ by SAS so ∠WQL ≅ ∠MLQ since congruent triangles have
congruent parts. So ML WQ since congruent alternate interior angles are formed by
parallel lines.
43. Yes
DB ≅ CE
∠BDC ≅ ∠ ECD
DC ≅DC
Reflexive
Δ BCD ≅ Δ EDC
SAS
44.
Not necessarily.
45.
Not necessarily.
Parent Guide with Extra Practice
85
CONVERSES
6.1.5
A conditional statement is a sentence in the “If – then” form. “If all sides are equal in
length, then a triangle is equilateral” is an example of a conditional statement. We can
abbreviate conditional statements by creating an arrow diagram. When the clause after
the “if” in a conditional statement exchanges places with the clause after the “then,” the
new statement is called the converse of the original. If the conditional statement is true,
the converse is not necessarily true, and vice versa.
See the Math Notes box in Lesson 6.1.5.
Example 1
Read each conditional statement below. Rewrite it as an arrow diagram, and state whether or not
it is true. Then write the converse of the statement, and state whether or not the converse is true.
a.
If a triangle is equilateral, then it is equiangular.
b.
If x = 4, then x 2 = 16 .
c.
If ABCD is a square, then ABCD is a parallelogram.
The arrow diagram for part (a) is not much shorter than the original statement:Δ is equilateral ⇒
Δ is equiangular
The converse is: If a triangle is equiangular, then it is equilateral. This statement and the original
conditional statement are both true.
In part (b), the conditional statement is true and the arrow diagram is: x = 4 ⇒ x 2 = 16 .
The converse of this statement, “If x 2 = 16 , then x = 4,” is not necessarily true because x could
equal –4.
In part (c), the arrow diagram is: ABCD is a square ⇒ ABCD is a parallelogram.
This statement is true, but the converse, “If ABCD is a parallelogram, then ABCD is a square,” is
not necessarily true. It could be a parallelogram or a rectangle.
86
Core Connections Geometry
Problems
Rewrite each conditional statement below as an arrow diagram and state whether or not it is true.
Then write the converse of the statement and state whether or not the converse is true.
1.
If an angle is a straight angle, then the angle measures 180°.
2.
If a triangle is a right triangle, then the sum of the squares of the lengths of the legs is equal
to the square of the length of the hypotenuse.
3.
If the measures of two angles of one triangle are equal to the measures of two angles of
another triangle, then the measures of the third angles are also equal.
4.
If one angle of a quadrilateral is a right angle, then the quadrilateral is a rectangle.
5.
If two angles of a triangle have equal measures, then the two sides of the triangle opposite
those angles have equal length.
Answers
1.
Conditional: True
180°
Converse: If an angle measures 180°, then it is a straight angle. True.
2.
Conditional: True
a
c
b
Converse: If the sum of the squares of the lengths of the legs is equal to the square of the
length of the hypotenuse, then the triangle is a right triangle. True.
3.
Conditional: True
Converse: If the measures of one pair of corresponding angles of two triangles are equal,
then the measures of the two other pairs of corresponding angles are also equal. False.
Parent Guide with Extra Practice
87
4.
C
Conditional: False
B
ABCD is a rectangle
A
D
Converse: If a quadrilateral is a rectangle, then one angle is a right angle. True, in fact,
all four angles are right angles.
5.
Conditional: True
Converse: If two sides of a triangle are equal in length, then the two angles opposite those
sides are equal in measure. True.
88
Core Connections Geometry
APPLICATIONS AND CONNECTIONS
6.2.1 – 6.2.5
The remaining sections of Chapter 6 are devoted to doing big problems. Students solve
problems that involve many of the topics that they have studied so far, giving them the
chance to connect the ideas and information as well as extend it to new situations.
Example 1
To frame a doorway, strips of wood surround the opening creating a “frame.”
If the doorway’s dimensions are 30 inches by 80 inches and the strips of
wood are 2 12 inches wide, how much wood is needed to frame the doorway
and how should it be cut? Assume that the strips will be assembled as shown
in the figure at right and that they are sold in 8’ lengths.
30
80
Cutting two pieces of wood 80 inches long, and one piece 30 inches long will not enable us to
make a frame for the doorway. The inside edges of the strips of wood will have those
measurements, but the outside dimensions of the wood are longer.
The wood strips will meet at a 45° angle at the corners of the doorframe. Looking at the corner
carefully (as shown at right) we see two 45°-45°-90° triangles in the corner. The lengths AD and
CD are both 2 12 inches, since they are the width of the strips of wood.
Since a 45°-45°-90° triangle is isosceles, this means AB and
B
C
BC are also 2 12 inches in length. Therefore, we need two
45°
strips that are 82 12 inches long ( 80 + 2 12 ), and one strip that is
45°
35 inches long ( 30 + 2 12 + 2 12 ), because each end must extend
the width of the vertical strip). Since the strips come in 8-foot
lengths, we would need to buy three of them. Two will be cut
at a 45° angle, with the outsides edge 82 12 inches long and the
A
D
inside edges 80 inches long. The third piece has two 45° angle
cuts. The outside length is 35 inches while the inside length is
30 inches.
Parent Guide with Extra Practice
89
Example 2
A friend offers to play a new game with
you, using the spinner shown at right.
Your friend says that you can choose to be
player 1 or 2. On each turn, you will spin
the spinner twice. If the letters are the
same, player 1 gets a point. If the letters
are different, then player 2 gets the point.
Which player would you choose to be?
Justify your answer.
A
120°
B
To help us decide which player to be, we will create an area model to represent the probabilities.
On the area model, the left edge represents the two outcomes from the first spin; the top edge
represents the outcomes from the second spin. On the first spin, there are two possible
outcomes, A and B, which are not equally likely. In fact, the probability of B occurring is
P(B) = 23 , while the probability of A occurring is P(A) = 13 . This is true for the first spin and the
second spin. We divide the area model according to these probabilities, and fill in the possible
outcomes.
Player 1 receives a point when the letters are the same for
both spins. This outcome is represented by the shaded
squares. Player 2 receives the point when the letters are
different. By multiplying the dimensions of each region,
the areas are expressed as ninths and we see that:
P(1 gets a point) =
1
9
+ 49 =
5
9
2
9
+ 29 =
4
9
and
P(2 gets a point) =
Since the probability that player 1 will win a point is greater than the probability that player 2
will, we should choose to be player 1.
90
Core Connections Geometry
Problems
1.
On graph paper, plot the points A(3, –4), B(8, –1), C(2, 9), and D(–3, 6) and connect them
in order. Find all the measurements of this shape (side lengths, perimeter, area, and angle
measures) and based on that information, decide the most specific name for this shape.
Justify your answer.
2.
The spinner at right is only partially
completed. Complete the spinner
based on these clues.
3
a. There are three other single
digit numbers on the spinner.
All four numbers on the spinner
are equally likely results for one
spin. No digit is repeated.
b. If the spinner is spun twice and
the two outcomes are added, the largest possible sum is 16, while the smallest
possible sum is 2. The most common sum is 9.
3.
To go along with the snowflakes you are making for the winter dance decorating
committee, you are going to make some “Star Polygons.” A Star Polygon is formed by
connecting equally spaced points on a circle in a specific order from a specified starting
point.
start
For instance, the circle at right has five equally spaced points.
If we connect them in order, the shape is a regular pentagon
(dashed sides). But if we connect every other point,
continuing until we reach the point that we started with, we
get a star.
a.
What happens when 6 points are equally spaced around a
circle? Under what conditions will you get a “normal”
polygon, and when will you get a “star polygon”?
b.
Explore other options. Come up with a rule that
explains when a normal polygon is formed when
connecting points, and when a star polygon is formed.
Consider various numbers of points.
Parent Guide with Extra Practice
start
91
Answers
1.
The side lengths are: AB = CD = 34 ≈ 5.83 units, AD = CB = 136 = 2 34 ≈ 11.66
units. Perimeter = 6 34 ≈ 34.99 units. The slope of AB = the slope of CD = 53 , slope of
AD = slope of CB = − 53 . Since the slopes are negative reciprocals, we know that the
segments are perpendicular, so all four angles are 90°, so the figure is a rectangle.
Area = 68 square units.
2.
The spinner is divided into four equal pieces, with the numbers 3, 1, 6, and 8.
3.
(a)
Connecting consecutive points forms a hexagon.
Connecting every other point forms an equilateral
triangle. Connecting every third point forms several
line segments (diameters), but no star.
(b)
Trying the same things with 7 points around a circle
produces more interesting results. Connecting each
point in order forms a heptagon (7-gon). Connecting
every second point or every fifth point produces the
first star polygon at right. Connecting every third point
or every fourth point produces the second star polygon
at right.
start
start
Every second point
or every fifth point
start
To generalize, if the number of points around the circle is n,
and we connect to the rth point, the polygon is a star polygon
if n and r have no common factors. Note that whenever r = 1
or r = n – 1, the result is always a polygon.
Every third point
or every fourth point
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Core Connections Geometry