# 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. | mA + mB = 90º , mC + mB = 90º | Definition of complementary |

3. | mA = 90 º - mB, mC = 90º - mB | Subtraction property of equality |

4. | mA = mC | 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 mABC = mABC - mDBC.

BD divides ABC into two angles, ABD and DBC.

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

Prove: mABD = mABC - mDBC.

ReasonsStatements | ||
---|---|---|

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

2. | mABD + mDBC = mABC | Angle Addition Postulate |

3. | mABD = mABC - mDBC | 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: mDBC = 0.

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

1. | BD and BE bisect ABC | Given |

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

3. | mABD = mDBC and mABE ~= mEBC | Definition of ~= |

4. | mABD + mDBE + mEBC = mABC | Angle Addition Postulate |

5. | mABD + mDBC = mABC and mABE + mEBC = mABC | Angle Addition Postulate |

6. | 2mABD = mABC and 2mEBC = mABC | Substitution (steps 3 and 5) |

7. | mABD = ^{mABC}/_{2} and mEBC = ^{mABC}/_{2} |
Algebra |

8. | ^{mABC}/_{2} + mDBE + ^{mABC}/_{2} = mABC |
Substitution (steps 4 and 7) |

9. | mABC + mDBE = mABC | Algebra |

10. | mDBE = 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. | mA + mB = 180º | Definition of supplementary angles |

3. | mA = 90º | Definition of right angle |

4. | 90º + mB = 180º | Substitution (steps 2 and 3) |

5. | mB = 90º | Algebra |

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

### Proving Relationships Between Lines

- m6 = 105º , m8 = 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 m1 + m2 = 180º | Definition of supplementary angles |

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

4. | 2 ~= 3 | Postulate 10.1 |

5. | m2 ~= m3 | Definition of ~= |

6. | m1 + m3 = 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.

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. | mB = 90º | Definition of right angle |

3. | mA + mB + mC = 180º | Theorem 11.1 |

4. | mA + 90º + mC = 180º | Substitution (steps 2 and 3) |

5. | mA + mC = 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.

ABC with exterior angle BCD.

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

1. | ABC with exterior angle BCD | Given |

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

3. | mBCA + mBCD = mDCA | Angle Addition Postulate |

4. | mBCA + mBCD = 180º | Substitution (steps 2 and 3) |

5. | mBAC + mABC + mBCA = 180º | Theorem 11.1 |

6. | mBAC + mABC + mBCA = mBCA + mBCD | Substitution (steps 4 and 5) |

7. | mBAC + mABC = mBCD | 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. | mABC = 90º and mDBC = 90º | Definition of right angles |

4. | mABC = mDBC | 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. | mABC = 90º and mDBC = 90º | Definition of right angles |

4. | mABC = mDBC | 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
- 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. | 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 = 43 , AB = 8 , RS = 16, RT = 83

3. AC = 42 , BC = 42

### 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. | mBMA = mDMA | Definition of ~= |

11. | MBD is a straight angle, and mBMD = 180º | Definition of straight angle |

12. | mBMA + mDMA = mBMD | Angle Addition Postulate |

13. | mBMA + mDMA = 180º | Substitution (steps 9 and 10) |

14. | 2mBMA = 180º | Substitution (steps 9 and 12) |

15. | mBMA = 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. | 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 ˆRST =
^{155}/_{18}feet - 9 feet
^{2} - 15 feet
^{2} - 28º

### The Unit Circle and Trigonometry

^{3}/_{34}=^{334}/_{34}^{1}/_{3}=^{3}/_{3}- tangent ratio =
^{40}/_{3}, sine ratio =^{40}/_{7} - tangent ratio =
^{5}/_{56}=^{556}/_{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.

To order this book direct from the publisher, visit the Penguin USA website or call 1-800-253-6476. You can also purchase this book at Amazon.com and Barnes & Noble.