# 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 = 90 , m?C + m?B = 90 | Definition of complementary |

3. | m?A = 90 - m?B, m?C = 90 - 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 = 180 | Definition of supplementary angles |

3. | m?A = 90 | Definition of right angle |

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

5. | m?B = 90 | Algebra |

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

### Proving Relationships Between Lines

- m?6 = 105 , m?8 = 75
- 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 = 180 | 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 = 180 | 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.

?ABC is a right triangle.

Given: ?ABC is a right triangle, and ?B is a right angle.

Prove: ?A and ?C are complementary angles.

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

1. | ?ABC is a right triangle, and ?B is a right angle | Given |

2. | m?B = 90 | Definition of right angle |

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

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

5. | m?A + m?C = 90 | 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.

?ABC with exterior angle ?BCD.

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

1. | ?ABC with exterior angle ?BCD | Given |

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

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

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

5. | m?BAC + m?ABC + m?BCA = 180 | 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: ?ABC ~= ?ABC.

Symmetric property: If ?ABC ~= ?DEF, then ?DEF ~= ?ABC.

Transitive property: If ?ABC ~= ?DEF and ?DEF ~= ?RST, then ?ABC ~= ?RST.

2. Proof: If AC ~= CD and ?ACB ~= ?DCB as shown in Figure 12.5, then ?ACB ~= ?DCB.

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

1. | AC ~= CD and ?ACB ~= ?DCB | Given |

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

3. | ?ACB ~= ?DCB | SAS Postulate |

3. If CB ? AD and ?ACB ~= ?DCB, as shown in Figure 12.8, then ?ACB ~= ?DCB.

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

1. | CB ? AD and ?ACB ~= ?DCB | Given |

2. | ?ABC and ?DBC are right angles | Definition of ? |

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

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

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

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

7. | ?ACB ~= ?DCB | ASA Postulate |

4. If CB ? AD and ?CAB ~= ?CDB, as shown in Figure 12.10, then ?ACB~= ?DCB.

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

1. | CB ? AD and ?CAB ~= ?CDB | Given |

2. | ?ABC and ?DBC are right angles | Definition of ? |

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

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

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

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

7. | ?ACB ~= ?DCB | AAS Theorem |

5. If CB ? AD and AC ~= CD, as shown in Figure 12.12, then ?ACB ~= ?DCB.

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

1. | CB ? AD and AC ~= CD | Given |

2. | ?ABC and ?DBC are right triangles | Definition of right triangle |

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

4. | ?ACB ~= ?DCB | 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. | ?PMN ~= RMQ | ASA Postulate |

6. | ?N ~= ?Q | CPOCTAC |

### Smiliar Triangles

- x = 11
- x = 12
- 40 and 140
- 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. | ?ACB ~ ?DCE | 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. | ?ABC ~ ?DBE | 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. | ?ABC ~= ?ADC | 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. | ?ABC ~= ?ADC | SSS Postulate |

5. | ?BAC ~= ?DAC | CPOCTAC |

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

7. | ?ABM ~= ?ADM | 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 = 180 | Definition of straight angle |

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

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

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

15. | m?BMA = 90 | 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. | ?ABC ~= ?CDA | 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: 20? feet, length of ?RST =
^{155}/_{18}? feet - 9? feet
^{2} - 15? feet
^{2} - 28

### 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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