Teacher Resource • Answer Key
Problems 5 – 8 + Exit Ticket
| # | Givens | Target Conclusion | Accepted Reason(s) |
|---|---|---|---|
| 5 | \(\overline{EF} \cong \overline{GH}\) | \(\overline{EG} \cong \overline{FH}\) | Segment Addition Postulate (Reflexive) |
| Adding common middle segment \(\overline{FG}\) (\(\overline{FG} \cong \overline{FG}\)). |
|
| 6 | \(M\) midpt \(\overline{AB}\), \(N\) midpt \(\overline{CD}\),
\(\overline{AM} \cong \overline{CN}\) | \(\overline{AB} \cong \overline{CD}\) | Doubles Postulate / Halves of Equals
Doubles of congruent segments are congruent.
| | 7 | \(\overline{RS} \cong \overline{TU}\), \(\overline{ST} \cong \overline{TU}\) | \(\overline{RS} \cong \overline{ST}\) (or \(\overline{RT} \cong \overline{SU}\)) | Transitive Property of Congruence
Or Segment Addition Postulate if stating \(RT \cong SU\).
| | 8 | \(\overline{KL} \cong \overline{YZ}\), \(\overline{LM} \cong \overline{XY}\) | \(\overline{KM} \cong \overline{XZ}\) | Segment Addition Postulate / Substitution
Addition Property of Congruence (\(KL + LM = YZ + XY\)).
|
Total: 10 Points
Question 1 (3 pts)
\(AC = 14.5 + 21.8 = \mathbf{36.3\text{ cm}}\)
Reason: Segment Addition Postulate
2 pts for correct calculation; 1 pt for stating Segment Addition Postulate.
Question 2 (3 pts)
\(FG = 42 - 27.5 = \mathbf{14.5\text{ in}}\)
Reason: Segment Subtraction Postulate
2 pts for correct calculation; 1 pt for Subtraction Postulate/Property.
Question 3 (4 pts)
Conclusion: \(\overline{RM} \cong \overline{MS}\) and \(RM = MS = \mathbf{18}\)
Reason: Halves Postulate
Also accept: Definition of Midpoint. 2 pts for conclusion; 2 pts for reason.
Geometry Proof Foundations • Unit 1 Page 2 of 2
Instructional Note: Highlight that the Halves Postulate states halves of congruent segments (or equal halves of a segment) are congruent.
Geometry Proof Foundations • Teacher Copy • Facilitation & Answer Key Reference
Conclusion: Remaining \(\angle\)s: Reason:
Geometry Flashcard Notes • Unit 1 Page 1 of 3
Geometry Flashcard Notes • Part 2
Page 2 of 3
3
Halves of Equals = Equal
Rule: Halves of congruent segments (or angles) are to each other. (Key keywords: midpoint divides a segment into halves; ray bisector divides an angle into halves!)
Card 5: Segment Halves Postulate
PMS ANT
Given: \(\overline{PS} \cong \overline{AT}\), \(M\) midpt \(\overline{PS}\), \(N\) midpt \(\overline{AT}\)
Conclusion (any half): \(\overline{PM} \cong \overline{AN}\) (or \(\overline{MS} \cong \overline{NT}\))
Reason: Halves Post
Card 6: Angle Halves Postulate
XYZK PCTM
Given: \(\vec{YK}\) bisects \(\angle XYZ\), \(\vec{CM}\) bisects \(\angle PCT\), \(\angle PCT \cong \angle XYZ\)
Conclusion: \(\angle XYK \cong \angle PCM\)
Reason: Halves Post
Your Turn: Segment Halves Practice
Given: \(\overline{AB} \cong \overline{CD}\), \(X\) is midpoint of \(\overline{AB}\), \(Y\) is midpoint of \(\overline{CD}\).
Conclusion: Reason:
4
Replace Equal with Equal
Rule: If two geometric quantities are both congruent to the quantity, one may be for the other!
Substitution with Angles
Given: \(\angle 7 \cong \angle 9\) & \(\angle 7 \cong \angle 3\)
Notice: Both are equal to \(\angle 7\)!
Conclusion: \(\angle 3 \cong \angle 9\)
Reason: Substitution Property
Substitution with Segments
Given: \(\overline{LM} \cong \overline{XY}\) & \(\overline{PQ} \cong \overline{XY}\)
Notice: Both are equal to \(\overline{XY}\)!
Conclusion: \(\overline{LM} \cong \overline{PQ}\)
Reason: Substitution Property
Your Turn: Substitution Practice
Given: \(\angle A \cong \angle B\) and \(\angle B \cong \angle C\).
Conclusion: Reason:
Geometry Flashcard Notes • Unit 1 Page 2 of 3
Geometry Flashcard Notes • Part 3
Page 3 of 3
5
Shared Middle Piece
Reflexive Property: Any geometric figure is congruent to (\(\overline{QR} \cong \overline{QR}\) or \(\angle 1 \cong \angle 1\)). When two figures share an overlapping middle piece, we use the Reflexive Property first, then add or subtract!
Card 12: Reflexive Segment Addition Add middle
OQRP
Given: \(\overline{OQ} \cong \overline{RP}\) (wings)
Step a: \(\overline{QR} \cong \overline{QR}\) (Reflexive)
Step b: \(\overline{OR} \cong \overline{QP}\) (Seg Add Post)
Card 11: Reflexive Segment Subtraction Subtract middle
OQRP
Given: \(\overline{OR} \cong \overline{QP}\) (overlapping wholes)
Step a: \(\overline{QR} \cong \overline{QR}\) (Reflexive)
Step b: \(\overline{OQ} \cong \overline{RP}\) (Seg Subtr Post)
Card 10: Reflexive Angle Addition Add middle \(\angle 1\)
617 STUVW
Given: \(\angle 6 \cong \angle 7\) (wings)
Step a: \(\angle 1 \cong \angle 1\) (Reflexive)
Step b: \(\angle STW \cong \angle VTU\) (Angle Add Post)
Card 9: Reflexive Angle Subtraction Subtract middle \(\angle 1\)
617
Given: \(\angle STW \cong \angle VTU\) (wholes)
Step a: \(\angle 1 \cong \angle 1\) (Reflexive)
Step b: \(\angle 6 \cong \angle 7\) (Angle Subtr Post)
| Reference Picture | Given Information | Conclusion | Postulate / Reason |
|---|---|---|---|
| AEB CFD | |||
| \(\overline{AB} \cong \overline{CD}\) |
\(\overline{AE} \cong \overline{FD}\)
| | | | ABCD DEFG |
\(\vec{BD}\) bisects \(\angle ABC\), \(\vec{EG}\) bisects \(\angle DEF\)
\(\angle ABC \cong \angle DEF\)
| | | | OQRP |
Points \(O-Q-R-P\) collinear
\(\overline{OQ} \cong \overline{RP}\)
| | | | 617 STU |
\(\angle STW \cong \angle VTU\)
(overlap \(\angle 1\))
| | |
Geometry Flashcard Notes • Unit 1 Page 3 of 3
Given: \(\angle PQR \cong \angle XYZ\) and \(\angle 1 \cong \angle 3\) (where \(\angle 1\) is inside \(\angle PQR\))
Conclusion: Remaining \(\angle\)s: \(\angle 2 \cong \angle 4\) Reason: Angle Subtr Post
Geometry Flashcard Notes Key • Unit 1 Page 1 of 3
Teacher Resource • Answer Key
Page 2 of 3
3
Halves of Equals = Equal
Rule: Halves of congruent segments (or angles) are congruent to each other. (Key keywords: midpoint divides a segment into halves; ray bisector divides an angle into halves!)
Card 5: Segment Halves Postulate
PMS ANT
Given: \(\overline{PS} \cong \overline{AT}\), \(M\) midpt \(\overline{PS}\), \(N\) midpt \(\overline{AT}\)
Conclusion (any half): \(\overline{PM} \cong \overline{AN}\) (or \(\overline{MS} \cong \overline{NT}\))
Reason: Halves Post
Card 6: Angle Halves Postulate
XYZK PCTM
Given: \(\vec{YK}\) bisects \(\angle XYZ\), \(\vec{CM}\) bisects \(\angle PCT\), \(\angle PCT \cong \angle XYZ\)
Conclusion: \(\angle XYK \cong \angle PCM\)
Reason: Halves Post
Your Turn Solution: Segment Halves
Given: \(\overline{AB} \cong \overline{CD}\), \(X\) is midpoint of \(\overline{AB}\), \(Y\) is midpoint of \(\overline{CD}\).
Conclusion: \(\overline{AX} \cong \overline{CY}\) (or \(\overline{XB} \cong \overline{YD}\)) Reason: Halves Postulate
4
Replace Equal with Equal
Rule: If two geometric quantities are both congruent to the same quantity, one may be substituted for the other!
Substitution with Angles
Given: \(\angle 7 \cong \angle 9\) & \(\angle 7 \cong \angle 3\)
Notice: Both are equal to \(\angle 7\)!
Conclusion: \(\angle 3 \cong \angle 9\)
Reason: Substitution Property
Substitution with Segments
Given: \(\overline{LM} \cong \overline{XY}\) & \(\overline{PQ} \cong \overline{XY}\)
Notice: Both are equal to \(\overline{XY}\)!
Conclusion: \(\overline{LM} \cong \overline{PQ}\)
Reason: Substitution Property
Your Turn Solution: Substitution
Given: \(\angle A \cong \angle B\) and \(\angle B \cong \angle C\).
Conclusion: \(\angle A \cong \angle C\) Reason: Substitution Property (or Transitive)
Geometry Flashcard Notes Key • Unit 1 Page 2 of 3
Teacher Resource • Answer Key
Page 3 of 3
5
Shared Middle Piece
Reflexive Property: Any geometric figure is congruent to itself (\(\overline{QR} \cong \overline{QR}\) or \(\angle 1 \cong \angle 1\)). When two figures share an overlapping middle piece, we use the Reflexive Property first, then add or subtract!
Card 12: Reflexive Segment Addition Add middle
OQRP
Given: \(\overline{OQ} \cong \overline{RP}\) (wings)
Step a: \(\overline{QR} \cong \overline{QR}\) (Reflexive)
Step b: \(\overline{OR} \cong \overline{QP}\) (Seg Add Post)
Card 11: Reflexive Segment Subtraction Subtract middle
OQRP
Given: \(\overline{OR} \cong \overline{QP}\) (overlapping wholes)
Step a: \(\overline{QR} \cong \overline{QR}\) (Reflexive)
Step b: \(\overline{OQ} \cong \overline{RP}\) (Seg Subtr Post)
Card 10: Reflexive Angle Addition Add middle \(\angle 1\)
617 STUVW
Given: \(\angle 6 \cong \angle 7\) (wings)
Step a: \(\angle 1 \cong \angle 1\) (Reflexive)
Step b: \(\angle STW \cong \angle VTU\) (Angle Add Post)
Card 9: Reflexive Angle Subtraction Subtract middle \(\angle 1\)
617
Given: \(\angle STW \cong \angle VTU\) (wholes)
Step a: \(\angle 1 \cong \angle 1\) (Reflexive)
Step b: \(\angle 6 \cong \angle 7\) (Angle Subtr Post)
| Reference Picture | Given Information | Target Conclusion | Postulate / Reason |
|---|---|---|---|
| AEB CFD | |||
| \(\overline{AB} \cong \overline{CD}\) |
\(\overline{AE} \cong \overline{FD}\)
| \(\overline{EB} \cong \overline{CF}\) | Segment Subtraction Postulate | | ABCD DEFG |
\(\vec{BD}\) bisects \(\angle ABC\), \(\vec{EG}\) bisects \(\angle DEF\)
\(\angle ABC \cong \angle DEF\)
| \(\angle ABD \cong \angle DEG\) | Halves Postulate | | OQRP |
Points \(O-Q-R-P\) collinear
\(\overline{OQ} \cong \overline{RP}\)
| \(\overline{OR} \cong \overline{QP}\) | Segment Addition Postulate (Reflexive) | | 617 STU |
\(\angle STW \cong \angle VTU\)
(overlap \(\angle 1\))
| \(\angle 6 \cong \angle 7\) | Angle Subtraction Postulate (Reflexive) |
Geometry Flashcard Notes Key • Unit 1 Page 3 of 3
Algebraic Segments • Unit 1 Page 2 of 2
| 4 | \(5x = 30\) | Addition Property of Equality |
| 5 | \(x = 6\) | Division Property of Equality |
10 Points Total
Question 1 (5 pts)
(4x - 5) + (3x + 2) = 46
7x - 3 = 46 → 7x = 49 → x = 7 (3 pts)
AB = 4(7) - 5 = 28 - 5 = 23 (2 pts)
Check: BC = 3(7)+2 = 23; 23+23 = 46.
Question 2 (5 pts)
6x - 11 = 2x + 13
4x = 24 → x = 6 (3 pts)
XM = 6(6)-11 = 25 → XY = 2(25) = 50 (2 pts)
Check: MY = 2(6)+13 = 25; XM = MY = 25.
Algebraic Segments Key • Unit 1 Page 2 of 2
