Download Unit 1b Study Guide

Geometry
Summative 2: Study Guide
Name_____________________________________________________________
State the Converse and Biconditional
1. If two angles add to 180 degrees, then the two angles are supplementary.
a. Converse:
b. Truth Value of Converse:
c. Biconditional (if possible):
2. If you are clapping your hands, then you are making a sound.
a. Converse:
b. Truth Value of Converse:
c. Biconditional (if possible):
Period_________
Justify the Conclusion:
3.
Definition of Congruence
Segment Addition Postulate
Angle Addition Postulate
Vertical Angles are Congruent
Definition of Supplementary
Definition of Complementary
Definition of a Midpoint
Reflexive Property
Given: B is the Midpoint of AC
4.
Conclusion: AB  BC
Conclusion: NM = TY
Why:
5.
Given: mL LMN + mLNMO = 90
Given: NM TY
Why:
6.
Given:
5
Conclusion:
Justification:
3
Conclusion:
Justification:
Finish the 2 Collumn Proof
7. (5 blanks)
Y
W
Given
Segment Addition Postulate
Prove
Substitution
Z
X
XY
Reflexive Property
WX
Given: WZ = 2YZ + XY
Subtraction
XZ
Definition of a Midpoint
Prove: YZ WX
1) WZ = 2YZ + XY
1)
2) WZ = WX + __ + YZ
2)
3) 2YZ + XY = WX + XY + YZ
3) Substitution
4) 2YZ + XY – YZ – XY=WX + XY + YZ – YZ – XY
4)
5) YZ = __
5) Simplify
6) YZ WX
6) Equal Lengths are Congruent
8. (5 blanks)
Given: mLD and mLE are Supplementary;
mLE and mLF are Supplementary
Prove: mLD mLF
Prove
Definition of Bisects
Given
Substitution
LF = LF
Definition of Supplementary
LE = LE
Equal Lengths are Congruent
1) mLD and mLE are Supplementary
mLE and mLF are Supplementary
2) mLD + mLE = 180; mLE+ mLF = 180
1)
3) mLD + mLE = mLE + mLF
3)
4)
4) Reflexive Property
5) mLD mLF
5) Subtraction
6) mLD  mLF
6)
2)
9. Complete this paragraph proof by filling in the missing words.
Given: M is the midpoint of ⃡𝐿𝑁, N is the midpoint of ⃡𝑀𝑂
̅̅̅̅ = ̅̅̅̅
Prove: 𝐿𝑀
𝑁𝑂
We are given that ______________________________________. We are also given that ___________
______________________________. By the definition of _________________ ̅̅̅̅
𝐿𝑀 = ̅̅̅̅̅
𝑀𝑁, and
̅̅̅̅̅
̅̅̅̅ = ̅̅̅̅
𝑀𝑁 = ̅̅̅̅
𝑁𝑂. We can then say that 𝐿𝑀
𝑁𝑂 by the _________________________________________.