Geo A
Unit 4 HW 13
p. 206 # 1 – 9, 11, 12, 14, 20
p. 264 # 3, 7, 9, 25
1.
1.
2.
3.
4.
5.
Statements
P is the midpt of XZ
XP = PZ
<1 = <2
WZ = WX
WY perpendicular bis XZ
McCaleb
Reasons
1. Given
2. Def midpt
3. Given
4.
5. Equidistant Thm
(2 equidistant pts  perpendicular bis.)
6.
XY = YZ
6. Equidistant Converse
(perpendicular bis  pt equidistant)
2.
a) <ABE and <ACD
b) <EBD and <BDC
3.
1.
2.
3.
4.
4.
1.
2.
3.
4.
5.
Statements
<ADB = <CDB
<ADB supp <CDB
<ADB & <CDB are rt <s
BD is an altitude
Statements
<1 = <4
FC bis <BFD
<2 = <3
<AFC = <EFC
<AFC supp <EFC
<AFC & <EFC are rt <s
1.
2.
3.
4.
Reasons
Given
Def linear pair
If 2 <s are = & supp, they are rt
Def altitude
Reasons
1. Given
2.
3.
4.
5.
Def bisect
< addition prop (<1 + <2 = <4 + <3)
Def linear pair
If 2 <s are = & supp, they are rt
A
5.
A
D
C
B
D
Given: ΔABC is isosceles with base BC
ΔDBC is isosceles with base BC
Prove: AD is the perpendicular bis of BC
1.
2.
3.
C
B
Statements
Reasons
ΔABC is isosceles with base BC 1. Given
ΔDBC is isosceles with base BC
AB = AC
2. Def isosceles
DB = DC
AD is the perpendicular bis of BC 3. Equidistant Thm
(2 equidistant pts  perpendicular bis.)
6.
1.
2.
3.
4.
Statements
ΔABC is isosceles w/ base AC
<1 = <2
AB = CB
AD = CD
BD perpendicular AC
Reasons
1. Given
2. Def isosceles
3.
4. Equidistant Thm
(2 equidistant pts  perpendicular bis.)
7.
1.
2.
3.
4.
5.
Statements
Circle O
M is the midpt of AB
MA = MB
Draw OA & OB
OA = OB
OM perpendicular to AB
Reasons
1. Given
2. Def midpt
3. 2 pts determine a line
4. All radii of a circle are =
5. Equidistant Thm
(2 equidistant pts  perpendicular bis.)
8.
A
P
B
O
C
R
D
Given: Circle O
AB = CD
P is the midpt of AB
R is the midpt of CD
Prove: OP = OR
9.
a) (midpoint—average!)
𝑦 −𝑦
b) m = 2 1
= 7–1
𝑥2 −𝑥1
c) m =
𝑦2 −𝑦1
𝑥2 −𝑥1
M(9, 4)
= 6
15 – 3
= 8–4 =
12
4
11 – 2
9
d) perpendicular to BC, so opp recip
e) From (2, 4) to (9, 4) = 9 – 2 = 7
11.
1.
2.
3.
4.
5.
Statements
Circle O
<1 = <2
XY = YW
Draw OX & OW
OX = OW
OY perpendicular to WX
=½
no! they are not // b/c they do not
have the same slope.
-2/1
Reasons
1. Given
2.
3. 2 pts determine a line
4. All radii of a circle are =
5. Equidistant Thm
(2 equidistant pts  perpendicular bis.)
12.
1.
2.
3.
4.
5.
6.
7.
14.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Statements
<1 = <2 = <3 = <4
BE = BF
BD = BD
ΔDBE = ΔDBF
<BED = <BFD
<BED supp <BEA
<BFD supp <BFC
<BEA = <BFC
ΔABE = ΔCBF
Reasons
1. Given
Statements
<WXY = <ZYX
WX = ZY
XY = XY
ΔWXY = ΔZYX
<ZXY = <WYX
<WXR = <ZYR
<WRX = <ZRY
<W = <Z
ΔWRX = ΔZRY
WR = RZ
Reasons
1. Given
2.
3.
4.
5.
Reflexive
SAS
CPCTC
Def linear pair
6. Angle subtraction post
7. ASA
2.
3.
4.
5.
6.
7.
8.
9.
Reflexive Prop
SAS
CPCTC
< subtraction Prop
Vert < thm
CPCTC
AAS or ASA
CPCTC
Another option for # 14:
14.
1.
2.
3.
4.
5.
6.
20.
1.
2.
3.
4.
5.
6.
7.
8.
9.
Statements
<WXY = <ZYX
WX = ZY
XY = XY
ΔWXY = ΔZYX
<ZXY = <WYX
XR = YR
WR = RZ
Reasons
1. Given
2.
3.
4.
5.
6.
Reflexive Prop
SAS
CPCTC
Segment Subtraction Prop
Statements
Reasons
AB = AF
1. Given
BC = FE
AE = AC
2. Segment addition prop
<A = <A
3. Reflexive
ΔACF = ΔAEB
4. SAS
<C = <E
5. CPCTC
<ABD = <AFD
<ABD supp <CBD
6. Def linear pair
<AFD supp <EFD
<CBD = <EFD
7. Supplements of = <s are =
ΔBCD = ΔFED
8. ASA
(or you could use the vertical <s and AAS)
CD = DE
9. CPCTC
p. 264
3.
(see p. 216, Thm 30)
x + 50 > 3x
50 > 2x
25 > x (or x < 25)
7.
a) yes, b/c SSI <s are supp
b) no! SSE <s would need to be supp, not congruent
c) 120 + x + 40 = 180
x = 20
3x = 3(20) = 60
yes, b/c alt int <s are congruent
x + 40 = 60
9.
1.
2.
3.
4.
5.
6.
7.
Statements
AG = BE
AB = CD
AC = BD
AG // BE
<GAB = <EBD
ΔGAC = ΔEBD
<ACG = <BDE
GC // ED
25.
140˚
50˚
20˚
140˚
40˚
50˚
20˚
20˚
140˚
40˚
30˚
50˚
20˚
20˚
140˚
40˚
30˚
30˚
50˚
20˚
20˚
x = 40 + 30 = 70˚
Reasons
1. Given
2.
3.
4.
5.
6.
7.
Segment addition prop
Given
// lines  corr <s =
SAS
CPCTC
Corr <s =  // lines