# Geometry: Answer Key

## Answer Key

This provides the answers and solutions for the Put Me in, Coach! exercise boxes, organized by sections.

### Taking the Burden out of Proofs

- Yes
- Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Given: ∠A and ∠B are complementary, and ∠C and ∠B are complementary.

Prove: ∠A ~= ∠C.

Statements | Reasons | |
---|---|---|

1. | ∠A and ∠B are complementary, and ∠C and ∠B are complementary. | Given |

2. | m∠A + m∠B = 90º , m∠C + m∠B = 90º | Definition of complementary |

3. | m∠A = 90 º - m∠B, m∠C = 90º - m∠B | Subtraction property of equality |

4. | m∠A = m∠C | Substitution (step 3) |

5. | ∠A ~= ∠C | Definition of ~= |

### Proving Segment and Angle Relationships

- If E is between D and F, then DE = DF − EF.

E is between D and F.

Given: E is between D and F

Prove: DE = DF − EF.

Statements | Reasons | |
---|---|---|

1. | E is between D and F | Given |

2. | D, E, and F are collinear points, and E is on ¯DF | Definition of between |

3. | DE + EF = DF | Segment Addition Postulate |

4. | DE = DF − EF | Subtraction property of equality |

2. If →BD divides ∠ABC into two angles, ∠ABD and ∠DBC, then m∠ABC = m∠ABC - m∠DBC.

→BD divides ∠ABC into two angles, ∠ABD and ∠DBC.

Given: →BD divides ∠ABC into two angles, ∠ABD and ∠DBC

Prove: m∠ABD = m∠ABC - m∠DBC.

Statements | Reasons | |
---|---|---|

1. | →BD divides ∠ABC into two angles, ∠ABD and ∠DBC | Given |

2. | m∠ABD + m∠DBC = m∠ABC | Angle Addition Postulate |

3. | m∠ABD = m∠ABC - m∠DBC | Subtraction property of equality |

3. The angle bisector of an angle is unique.

∠ABC with two angle bisectors: →BD and →BE.

Given: ∠ABC with two angle bisectors: →BD and →BE.

Prove: m∠DBC = 0.

Statements | Reasons | |
---|---|---|

1. | →BD and →BE bisect ∠ABC | Given |

2. | ∠ABC ~= ∠DBC and ∠ABE ~= ∠EBC | Definition of angel bisector |

3. | m∠ABD = m∠DBC and m∠ABE ~= m∠EBC | Definition of ~= |

4. | m∠ABD + m∠DBE + m∠EBC = m∠ABC | Angle Addition Postulate |

5. | m∠ABD + m∠DBC = m∠ABC and m∠ABE + m∠EBC = m∠ABC | Angle Addition Postulate |

6. | 2m∠ABD = m∠ABC and 2m∠EBC = m∠ABC | Substitution (steps 3 and 5) |

7. | m∠ABD = ^{m∠ABC}/_{2} and m∠EBC = ^{m∠ABC}/_{2} | Algebra |

8. | ^{m∠ABC}/_{2} + m∠DBE + ^{m∠ABC}/_{2} = m∠ABC | Substitution (steps 4 and 7) |

9. | m∠ABC + m∠DBE = m∠ABC | Algebra |

10. | m∠DBE = 0 | Subtraction property of equality |

4. The supplement of a right angle is a right angle.

∠A and ∠B are supplementary angles, and ∠A is a right angle.

Given: ∠A and ∠B are supplementary angles, and ∠A is a right angle.

Prove: ∠B is a right angle.

Statements | Reasons | |
---|---|---|

1. | ∠A and ∠B are supplementary angles, and ∠A is a right angle | Given |

2. | m∠A + m∠B = 180º | Definition of supplementary angles |

3. | m∠A = 90º | Definition of right angle |

4. | 90º + m∠B = 180º | Substitution (steps 2 and 3) |

5. | m∠B = 90º | Algebra |

6. | ∠B is a right angle | Definition of right angle |

### Proving Relationships Between Lines

- m∠6 = 105º , m∠8 = 75º
- Theorem 10.3: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: ∠1 ~= ∠3.

Statements | Reasons | |
---|---|---|

1. | l m cut by a transversal t | Given |

2. | ∠1 and ∠2 are vertical angles | Definition of vertical angles |

3. | ∠2 and ∠3 are corresponding angles | Definition of corresponding angles |

4. | ∠2 ~= ∠3 | Postulate 10.1 |

5. | ∠1 ~= ∠2 | Theorem 8.1 |

6. | ∠1 ~= ∠3 | Transitive property of 3. |

3. Theorem 10.5: If two parallel lines are cut by a transversal, then the exterior angles on the same side of the transversal are supplementary angles.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: ∠1 and ∠3 are supplementary.

Statement | Reasons | |
---|---|---|

1. | l m cut by a transversal t | Given |

2. | ∠1 and ∠2 are supplementary angles, and m∠1 + m∠2 = 180º | Definition of supplementary angles |

3. | ∠2 and ∠3 are corresponding angles | Definition of corresponding angles |

4. | ∠2 ~= ∠3 | Postulate 10.1 |

5. | m∠2 ~= m∠3 | Definition of ~= |

6. | m∠1 + m∠3 = 180º | Substitution (steps 2 and 5) |

7. | ∠1 and ∠3 are supplementary | Definition of supplementary |

4. Theorem 10.9: If two lines are cut by a transversal so that the alternate exterior angles are congruent, then these lines are parallel.

Lines l and m are cut by a transversal t.

Given: Lines l and m are cut by a transversal t, with ∠1 ~= ∠3.

Prove: l m.

Statement | Reasons | |
---|---|---|

1. | Lines l and m are cut by a transversal t, with ∠1 ~= ∠3 | Given |

2. | ∠1 and ∠2 are vertical angles | Definition of vertical angles |

3. | ∠1 ~= ∠2 | Theorem 8.1 |

4. | ∠2 ~= ∠3 | Transitive property of ~=. |

5. | ∠2 and ∠3 are corresponding angles | Definition of corresponding angles |

6. | l m | Theorem 10.7 |

5. Theorem 10.11: If two lines are cut by a transversal so that the exterior angles on the same side of the transversal are supplementary, then these lines are parallel.

Lines l and m are cut by a t transversal t.

Given: Lines l and m are cut by a transversal t, ∠1 and ∠3 are supplementary angles.

Prove: l m.

Statement | Reasons | |
---|---|---|

1. | Lines l and m are cut by a transversal t, and ∠1 are ∠3 supplementary angles | Given |

2. | ∠2 and ∠1 are supplementary angles | Definition of supplementary angles |

3. | ∠3 ~= ∠2 | Example 2 |

4. | ∠3 and ∠2 are corresponding angles | Definition of corresponding angles |

5. | l m | Theorem 10.7 |

### Two's Company. Three's a Triangle

- An isosceles obtuse triangle
- The acute angles of a right triangle are complementary.

ΔABC is a right triangle.

Given: ΔABC is a right triangle, and ∠B is a right angle.

Prove: ∠A and ∠C are complementary angles.

Statement | Reasons | |
---|---|---|

1. | ΔABC is a right triangle, and ∠B is a right angle | Given |

2. | m∠B = 90º | Definition of right angle |

3. | m∠A + m∠B + m∠C = 180º | Theorem 11.1 |

4. | m∠A + 90º + m∠C = 180º | Substitution (steps 2 and 3) |

5. | m∠A + m∠C = 90º | Algebra |

6. | ∠A and ∠C are complementary angles | Definition of complementary angles |

3. Theorem 11.3: The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.

ΔABC with exterior angle ∠BCD.

Statement | Reasons | |
---|---|---|

1. | ΔABC with exterior angle ∠BCD | Given |

2. | ∠DCA is a straight angle, and m∠DCA = 180º | Definition of straight angle |

3. | m∠BCA + m∠BCD = m∠DCA | Angle Addition Postulate |

4. | m∠BCA + m∠BCD = 180º | Substitution (steps 2 and 3) |

5. | m∠BAC + m∠ABC + m∠BCA = 180º | Theorem 11.1 |

6. | m∠BAC + m∠ABC + m∠BCA = m∠BCA + m∠BCD | Substitution (steps 4 and 5) |

7. | m∠BAC + m∠ABC = m∠BCD | Subtraction property of equality |

4. 12 units^{2}

5. 30 units^{2}

6. No, a triangle with these side lengths would violate the triangle inequality.

### Congruent Triangles

1. Reflexive property: ΔABC ~= ΔABC.

Symmetric property: If ΔABC ~= ΔDEF, then ΔDEF ~= ΔABC.

Transitive property: If ΔABC ~= ΔDEF and ΔDEF ~= ΔRST, then ΔABC ~= ΔRST.

2. Proof: If ¯AC ~= ¯CD and ∠ACB ~= ∠DCB as shown in Figure 12.5, then ΔACB ~= ΔDCB.

Statement | Reasons | |
---|---|---|

1. | ¯AC ~= ¯CD and ∠ACB ~= ∠DCB | Given |

2. | ¯BC ~= ¯BC | Reflexive property of ~= |

3. | ΔACB ~= ΔDCB | SAS Postulate |

3. If ¯CB ⊥ ¯AD and ∠ACB ~= ∠DCB, as shown in Figure 12.8, then ΔACB ~= ΔDCB.

Statement | Reasons | |
---|---|---|

1. | ¯CB ⊥ ¯AD and ∠ACB ~= ∠DCB | Given |

2. | ∠ABC and ∠DBC are right angles | Definition of ⊥ |

3. | m∠ABC = 90º and m∠DBC = 90º | Definition of right angles |

4. | m∠ABC = m∠DBC | Substitution (step 3) |

5. | ∠ABC ~= ∠DBC | Definition of ~= |

6. | ¯BC ~= ¯BC | Reflexive property of ~= |

7. | ΔACB ~= ΔDCB | ASA Postulate |

4. If ¯CB ⊥ ¯AD and ∠CAB ~= ∠CDB, as shown in Figure 12.10, then ΔACB~= ΔDCB.

Statement | Reasons | |
---|---|---|

1. | ¯CB ⊥ ¯AD and ∠CAB ~= ∠CDB | Given |

2. | ∠ABC and ∠DBC are right angles | Definition of ⊥ |

3. | m∠ABC = 90º and m∠DBC = 90º | Definition of right angles |

4. | m∠ABC = m∠DBC | Substitution (step 3) |

5. | ∠ABC ~= ∠DBC | Definition of ~= |

6. | ¯BC ~= ¯BC | Reflexive property of ~= |

7. | ΔACB ~= ΔDCB | AAS Theorem |

5. If ¯CB ⊥ ¯AD and ¯AC ~= ¯CD, as shown in Figure 12.12, then ΔACB ~= ΔDCB.

Statement | Reasons | |
---|---|---|

1. | ¯CB ⊥ ¯AD and ¯AC ~= ¯CD | Given |

2. | ΔABC and ΔDBC are right triangles | Definition of right triangle |

3. | ¯BC ~= ¯BC | Reflexive property of ~= |

4. | ΔACB ~= ΔDCB | HL Theorem for right triangles |

6. If ∠P ~= ∠R and M is the midpoint of ¯PR, as shown in Figure 12.17, then ∠N ~= ∠Q.

Statement | Reasons | |
---|---|---|

1. | ∠P ~= ∠R and M is the midpoint of ¯PR | Given |

2. | ¯PM ~= ¯MR | Definition of midpoint |

3. | ∠NMP and ∠RMQ are vertical angles | Definition of vertical angles |

4. | ∠NMP ~= ∠RMQ | Theorem 8.1 |

5. | ΔPMN ~= RMQ | ASA Postulate |

6. | ∠N ~= ∠Q | CPOCTAC |

### Smiliar Triangles

- x = 11
- x = 12
- 40º and 140º
- If ∠A ~= ∠D as shown in Figure 13.6, then
^{BC}/_{AB}=^{CE}/_{DE}.

Statement | Reasons | |
---|---|---|

1. | ∠A ~= ∠D | Given |

2. | ∠BCA and ∠DCE are vertical angles | Definition of vertical angles |

3. | ∠BCA ~= ∠DCE | Theorem 8.1 |

4. | ΔACB ~ ΔDCE | AA Similarity Theorem |

5. | ^{BC}/_{AB} = ^{CE}/_{DE} | CSSTAP |

5. 150 feet.

### Opening Doors with Similar Triangles

- If a line is parallel to one side of a triangle and passes through the midpoint of a second side, then it will pass through the midpoint of the third side.

¯DE ¯AC and D is the midpoint of ¯AB.

Given: ¯DE ¯AC and D is the midpoint of ¯AB.

Prove: E is the midpoint of ¯BC.

Statement | Reasons | |
---|---|---|

1. | ¯DE ¯AC and D is the midpoint of ¯AB. | Given |

2. | ¯DE ¯AC and is cut by transversal ↔AB | Definition of transversal |

3. | ∠BDE and ∠BAC are corresponding angles | Definition of corresponding angles |

4. | ∠BDE ~= ∠BAC | Postulate 10.1 |

5. | ∠B ~= ∠B | Reflexive property of ~= |

6. | ΔABC ~ ΔDBE | AA Similarity Theorem |

7. | ^{DB}/_{AB} = ^{BE}/_{BC} | CSSTAP |

8. | DB = ^{AB}/_{2} | Theorem 9.1 |

9. | ^{DB}/_{AB} = ^{1}/_{2} | Algebra |

10. | ^{1}/_{2} = ^{BE}/_{BC} | Substitution (steps 7 and 9) |

11. | BC = 2BE | Algebra |

12. | BE + EC = BC | Segment Addition Postulate |

13. | BE + EC = 2BE | Substitution (steps 11 and 12) |

14. | EC = BE | Algebra |

15. | E is the midpoint of ¯BC | Definition of midpoint |

2. AC = 4√3 , AB = 8√ , RS = 16, RT = 8√3

3. AC = 4√2 , BC = 4√2

### Putting Quadrilaterals in the Forefront

- AD = 63, BC = 27, RS = 45
- ¯AX, ¯CZ, and ¯DY

Trapezoid ABCD with its XB CY four altitudes shown.

3. Theorem 15.5: In a kite, one pair of opposite angles is congruent.

Kite ABCD.

Given: Kite ABCD.

Prove: ∠B ~= ∠D.

Statement | Reasons | |
---|---|---|

1. | ABCD is a kite | Given |

2. | ¯AB ~= ¯AD and ¯BC ~= ¯DC | Definition of a kite |

3. | ¯AC ~= ¯AC | Reflexive property of ~= |

4. | ΔABC ~= ΔADC | SSS Postulate |

5. | ∠B ~= ∠D | CPOCTAC |

4. Theorem 15.6: The diagonals of a kite are perpendicular, and the diagonal opposite the congruent angles bisects the other diagonal.

Kite ABCD.

Given: Kite ABCD.

Prove: ¯BD ⊥ ¯AC and ¯BM ~= ¯MD.

Statement | Reasons | |
---|---|---|

1. | ABCD is a kite | Given |

2. | ¯AB ~= ¯AD and ¯BC ~= ¯DC | Definition of a kite |

3. | ¯AC ~= ¯AC | Reflexive property of ~= |

4. | ΔABC ~= ΔADC | SSS Postulate |

5. | ∠BAC ~= ∠DAC | CPOCTAC |

6. | ¯AM ~= ¯AM | Reflexive property of ~= |

7. | ΔABM ~= ΔADM | SAS Postulate |

8. | ¯BM ~= ¯MD | CPOCTAC |

9. | ∠BMA ~= ∠DMA | CPOCTAC |

10. | m∠BMA = m∠DMA | Definition of ~= |

11. | ∠MBD is a straight angle, and m∠BMD = 180º | Definition of straight angle |

12. | m∠BMA + m∠DMA = m∠BMD | Angle Addition Postulate |

13. | m∠BMA + m∠DMA = 180º | Substitution (steps 9 and 10) |

14. | 2m∠BMA = 180º | Substitution (steps 9 and 12) |

15. | m∠BMA = 90º | Algebra |

16. | ∠BMA is a right angle | Definition of right angle |

17. | ¯BD ⊥ ¯AC | Definition of ⊥ |

5. Theorem 15.9: Opposite angles of a parallelogram are congruent.

Parallelogram ABCD.

Given: Parallelogram ABCD.

Prove: ∠ABC ~= ∠ADC.

Statement | Reasons | |
---|---|---|

1. | Parallelogram ABCD has diagonal ¯AC. | Given |

2. | ΔABC ~= ΔCDA | Theorem 15.7 |

3. | ∠ABC ~= ∠ADC | CPOCTAC |

6. 144 units^{2}

7. 180 units^{2}

8. Kite ABCD has area 48 units^{2}.

Parallelogram ABCD has area 150 units^{2}.

Rectangle ABCD has area 104 units^{2}.

Rhombus ABCD has area ^{35}/_{2} units^{2}.

### Anatomy of a Circle

- Circumference: 20π feet, length of ˆRST =
^{155}/_{18}π feet - 9π feet
^{2} - 15π feet
^{2} - 28º

### The Unit Circle and Trigonometry

^{3}/_{√34}=^{3√34}/_{34}^{1}/_{√3}=^{√3}/_{3}- tangent ratio =
^{√40}/_{3}, sine ratio =^{√40}/_{7} - tangent ratio =
^{5}/_{√56}=^{5√56}/_{56}, cosine ratio =^{√56}/_{9}

Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with **Alpha Books**, a member of Penguin Group (USA) Inc.

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