Unit 3 HW 11
p. 158 # 2 – 5, 8, 10, 13, 17, 18
2.
Mrs. McCaleb
Statements
MO perpendicular OP
RP perpendicular OP
<MOP is a right <
<RPO is a right <
MP = RO
OP = PO
ΔMOP = ΔRPO
Reasons
1. Given
Statements
Circle O
OY = OZ
YO perpendicular YX
ZO perpendicular ZX
<OYX is right
<OZY is right
Draw OX
OX = OX
ΔYOX = ΔZOX
YX = ZX
Reasons
1. Given
2. All radii of a circle are =
3. Given
Reasons
1. Given
2.
3.
Statements
<BFA is a right <
<DEC is a right <
AB = CD
AE = CF
AF = CE
4.
5.
ΔDEC = ΔBFA
<CDE = <ABF
1.
2.
3.
4.
5.
3.
1.
2.
3.
4.
5.
6.
7.
8.
4.
1.
2. Def perpendicular
3. Given
4. Reflexive prop
5. HL
4. Def perpendicular
5.
6.
7.
8.
2 pts determine a line
Reflexive
HL
CPCTC
2. Given
3. Add a segment to 2 congruent
segments, you get 2 congruent segments
4. HL
5. CPCTC
5. To set up your own proof, you need givens to be 1) an isosceles Δ; and 2) the
altitude drawn to the base. Your “prove” statement should be that the 2 Δs are
J
congruent.
Given: ΔJAM is isosceles with JA = JM
JB is an altitude
Prove: ΔJAB = ΔJMB
2
A
Statements
JA = JM
JB is an altitude
<s 1 & 2 are right
JB = JB
ΔJAB = ΔJMB
Reasons
1. Given
Reasons
1. Given
4.
Statements
BD perpendicular CF
GE perpendicular CF
<BDC is a right <
<GEF is a right <
BC = GF
CE = DF
CD = FE
5.
6.
7.
ΔBDC = ΔGEF
<C = <F
ΔACF is isosceles
1.
2.
3.
4.
8.
1.
2.
3.
1
B
M
2. Def altitude
3. Reflexive
4. HL
2. Def perpendicular
3. Given
4. Subtract a segment from congruent
segments, you get congruent segments
5. HL
6. CPCTC
7. Def isosceles (it has 2 = <s)
10.
1.
2.
3.
4.
5.
6.
7.
13.
1.
2.
3.
4.
5.
6.
7.
17.
1.
2.
3.
4.
5.
6.
7.
8.
Statements
Circle P
SP = VP
ST = VT
Draw PT
PT = PT
ΔPST = ΔPVT
<PST = <PVT
1.
2.
3.
4.
5.
6.
7.
Reasons
Given
All radii of a circle are =
Given
2 pts determine a line
Reflexive
SSS
CPCTC
Statements
AB perpendicular BC
AB perpendicular BD
<ABC is a right <
<ABD is a right <
AC = AD
AB = AB
ΔABC = ΔABD
BC = BD
If CD is drawn, ΔBCD
will be isosceles
Reasons
1. Given
Statements
<R & <W are right
RX = WX
Draw XT
XT = XT
ΔRXT = ΔWXT
RT = WT
VT is 4/7 of WT
RS is 3/7 of RT
ST is 4/7 of RT
ST = TV
Reasons
1. Given
2. Def perpendicular
3.
4.
5.
6.
7.
2.
3.
4.
5.
6.
Given
Reflexive prop
HL
CPCTC
Def isosceles (2 = sides)
2 pts determine a line
Reflexive
HL
CPCTC
Given
7. Subtraction (1 – 3/7 = 4/7)
8. Like multiples of = segments are
= (multiply both by 4/7)
18. a) A is = E by ASA, so by CPCTC A gets the circle mark, and E gets the one
tick mark. Now B is = to either A or E by HL & C is = to either A or E by SSS.
There is no way to get D congruent, so
A, B, C, E
b) list all possibilities
A&B
A&C
A&D
A&E
B&C
B&D
B&E
C&D
C&E
D&E
10 possibilities. Which have congruent triangles?
A&B
A&C
A&D
A&E
B&C
B&D
B&E
C&D
C&E
D&E
The probability of getting congruent triangles is 6/10 = 3/5