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Transcript
```G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
1
Proof of geometric relationships continues in this objective. We establish some very important relationships
dealing with triangles and their angles.
Prove that the interior angles of a triangle sum to 180.
Informal Proof - Triangle Dissection
An informal proof that is often used is the process of having our students create a triangle on a piece of paper,
naming the three angles A, B, and C and then cutting out the triangle. When the triangle is cut out, the
student should rip off the three angles, placing them together, vertex to vertex. They will see that the three
angles form a straight line. Therefore, the sum of the three interior angles of a triangle is 180.
Classic Proof – Use of Parallel Line Angle Relationships
A formal two column proof of this theorem is done using the angle relationships found with parallel lines and a
transversal. Since these angle relationships were established in objective G.CO.9, we are now able to apply
them.
Given: ABC
B
2
Prove: m1 + m2 + m3 = 180

Construct an auxiliary line parallel to AC through B.
STATEMENT
AC || BD
m4 + m2 + m5 = 180
m1 = m4
m3 = m5
m1 + m2 + m3 = 180
REASON
Given (Auxiliary Line)
Angles of a Straight Angle
If , Alternate Interior ’s 
If , Alternate Interior ’s 
Substitution Property (Twice)
1
3
A
D
C
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
2
Transformational Proof
Given: ABC
B
2
Prove: m1 + m2 + m3 = 180
A
Translate ABC by vector AB to form straight ABB’
along the vector. The isometric properties of translation
preserve angles, thus m1 = mB’BC’. Since ABB’ is a
straight angle, we know that m2 + mCBC’ + mB’BC’
= 180Translations also preserve parallelism, therefore
BC ' . Since AC
BC ' m3 =
ensuring that AC
mCBC’ because alternate interior angles are congruent.
By making two substitutions into the straight angle
relationship of m2 + mCBC’ + mB’BC’ = 180 we
arrive at the proof that m1 + m2 + m3 = 180

Determine mA.
1
C
3
B'
1
B =A'
A
C'
3
2
1
C
3
A triangle has 180 as its angle sum…. So…
B
84°
180 - 84 - 35 = 61
35°
mA = 61
C
A
NYTS
Determine the measure of the angle.
a) mC = ________
b) mB = ________
c) mA = ________
B
B
d) mA = ________
B
B
o
96°
C
32°
44°
61°
A
A
60°
A
C
C
o
A
180 – 96 - 44 = 40
40 = mC
180 – 61 - 60 = 59
59 = mB
180 – 90 - 32 = 58
58 = mA
50°
C
180 – 50 = 130
130 / 2 = 65 = mA
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
3
Prove that Base Angles of an Isosceles are Equal (Isosceles Triangle Theorem)
Informal Proof - Paper Folding
Prove that the base angles of an isosceles triangle are congruent.
Create an isosceles triangle by using your compass to construct a
are congruent) and connect the endpoints with a
segment. Label the radii, AB and CB . Then fold the
paper until point A maps to point C. Crease the paper.
Notice that when you do this A  C. Therefore, the
base angles of an isosceles triangle are congruent.
B
C
D
A
Classic Proof – Triangle Congruence
Given: ABC is an isosceles triangle, with base AC .
Prove: A  C
Construct an auxiliary line that is a perpendicular bisector of AC . (B is
B
on the perpendicular bisector of AC because AB  BC )
STATEMENT
ABC is an Isosceles.
BD is the  bisector of B
BD  BD
ABD  CBD
A  C
REASON
Given
Given (Auxiliary Line)
Definition of a  Bisector
(D is the midpoint of AC )
Definition of a  Bisector
( BD is  to AC )
All Right Angles are 
Reflexive Property (Common Side)
A
D
SAS
CPCTC (Corresponding Parts of
Congruent Triangles are Congruent)
Alternative methods to prove congruent triangles: HL ( AB  CB, Definition of Isosceles Triangle) or SSS (
AB  CB, Definition of Isosceles Triangle and BD  BD , Reflexive Property)
C
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
4
Transformational Proof
Given: ABC is an isosceles triangle, with base AC .
B
Prove: A  C
Construct an auxiliary line that is a perpendicular bisector of AC .
By the definition of isosceles triangle, AB  CB. Because point B is
equidistant to points A and C, B is on the perpendicular bisector of AC .
A reflection over the perpendicular bisector would map A onto C, B onto
B, and D onto D. Thus the isometric properties of a reflection then give
us ABD  CBD. Therefore, A  C by corresponding parts of
congruent triangles are congruent.
C
D
A
Symmetry Proof
Given: ABC is an isosceles triangle, with base AC .
B
Prove: A  C
Construct an auxiliary line; BD , such that BD is the perpendicular
bisector of AC. Established in G.CO.3, an isosceles triangle has
reflectional symmetry about the perpendicular bisector of its base.
Thus A  C because A reflects onto C.
Determine mB.
D
A
C
A triangle has 180 as its angle sum…. So…
A
180 - 54 = 126
54°
126 / 2 = 63

mB = 63
C
B
Determine the measure of the requested angle.
a) mC = ________
b) mA = ________
A
A
c) mC = ________
d) mD = ________
A
A
98°
C
78°
B
B
36°
C
C
65°
B
D
B
C
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
180 – 98 = 82
130 / 2 = 41 = mC
180 – 78 -78 = 24
24 = mA
5
65 = mC
180 – 36 = 144
144 / 2 = 72 = mACB
180 – 72 = 108
180 – 108 = 72
72 / 2 = 36 = mD
Prove that the external angle of a triangle is equal to the sum of the two internal
remote angles. (Exterior Angle Theorem)
A
First of all we need to define these new terms.
An external angle of a polygon is the angle formed between a side’s
extension and its adjacent side. No part of this angle exists in the
interior of the triangle. (ACD is an external angle.)
D
C
B
A
The remote angles of a triangle are those angles that are not the
linear pair with the external angle being referenced to.
o
(A and B are remote angles to external ACD.)
x
B
D
C
Informal Proof – Angle Dissection
A
Prove that the external angle of a triangle is equal to
the sum of the two internal remote angles.
An informal proof that is often used is the process of
having our students create a triangle on a piece of
paper, naming the three angles A, B, and C. Cut out the
triangle and then draw a line on a piece of paper and
then lie the triangle down on the line so that one side is
extended by it. The student should rip off the two
remote angles and then place them together, vertex to
vertex in the external angle. They will see that the two
remote angles exactly fit into the external angle.
Therefore, the external angle of a triangle is equal to
the sum of the two internal remote angles.
B
C
D
C
D
Classic Proof – Triangle Congruence
Given: ABC with external angle, ACD.
A
Prove: mACD = mB + mA
B
C
D
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
STATEMENT
mB + mA + mACB = 180
mACB + mACD = 180
mACB + mACD = mB + mA + mACB
mACD = mB + mA
6
REASON
 sum 180
Linear Pairs are supplementary
Substitution Property (or Transitive)
+/- Property of Equality
Transformational Proof
A
Given: ABC with external angle, ACD.
Prove: mACD = mB + mA
B
C
D
A
Translate ABC by vector BC placing ABC in the
interior of external ACD. One property of a
translation is that it produces parallel lines and so
AB || AB '|| AC . Thus BAC  A’CA because
alternate interior angle are equal (or a rotation of
180 about the midpoint of AC )
A'
B
D
C = B'
C = B''
A
A'
B
Determine mBCD.
C = B'
D
The external BCD is equal to the sum of
the two remotes, A and B.
B
92°
46°
C
D
mBCD = 92 + 46
mBCD = 138
A
Determine the missing information.
a) mBCD = ___________
C = B''
b) mB = _________
c) mA = _________
mBCA = _________
mBCA = _________
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
B
7
B
B
42°
73°
74°
C
A
D
21°
D
C
65°
C
160°
A
D
A
73 + 74 = 147 = mBCD
65 - 21 = 44 = mB
180 - 65 = 115 = mBCA
160 - 42 = 118 = mA
180 - 160 = 20 = mBCA
Prove the Midsegment Theorem (that the segment joining midpoints of two
sides of a triangle is parallel to the third side and half the length)
Informal Proof
B
This isn’t necessarily proving the relationship but
we are going to work backwards to show that it
works.
Translate ABC by vector AB resulting in BB’D.
Translate ABC by vector AC resulting in CDE.
2
1
C
B'
These two transformations create AB’E, where B is
the midpoint of AB ' and D is the midpoint of B ' E ,
D
B
thus BD is the midsegment.
BD || AE Because of the translation by vector
1
AB and BD = AE because of the translation of AC
2
to BD and to CE.
3
A
2
1
3
A
E
C
Transformational Proof - Translation
Given: ABC, where D is the midpoint of AB and E is the
B
midpoint of BC .
1
Prove: DE || AC and DE = AC
2
D
A
E
C
G.CO.C.10 STUDENT NOTES & PRACTICE WS #1 – geometrycommoncore.com
8
B
Translate DBE by vector DA . If the translation of E by vector DA
lies on AC , then DE || AC by the properties of a translation.
Using an indirect proof, we assume that L (the image of E) does
not lie on AC. A contradiction occurs (provided below), so L the
image of E and lies on AC. In addition, by the properties of
E
D
A
C
L
translations, DE || AC .
Translate DBE by vector EC (vector EC is the same as vector DL because they are parallel and
the same distance). We know that DE = AL by the previous translation, and we know the DE = LC by
the current translation, therefore, AL = LC (transitive). Thus, AL + LC = AC (Segment Addition
Postulate), DE + DE = AC (Substitution), and 2DE = AC (Simplify). Lastly, by the Multiplication /
1
Division Property, DE  AC .
2
Determine x.
The midsegment theorem tells us that the
midsegment is half its parallel side.
x
½ (24 cm) = x
24 cm
12 cm = x
Determine the missing information.
a) x = ____________
b) x = ____________
c) x = ____________
68°
13.5 cm
x
x = 27 cm
3x
x
42 cm
3x = 21
x=7
x = 68
```
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