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

### Taking the Burden out of Proofs

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

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

Prove: ∠A ~= ∠C.

StatementsReasons
1.∠A and ∠B are complementary, and ∠C and ∠B are complementary.Given
2.m∠A + m∠B = 90º , m∠C + m∠B = 90ºDefinition 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

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

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.

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.

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

Prove: m∠DBC = 0.

StatementsReasons
1. →BD and →BE bisect ∠ABCGiven
2.∠ABC ~= ∠DBC and ∠ABE ~= ∠EBCDefinition of angel bisector
3.m∠ABD = m∠DBC and m∠ABE ~= m∠EBCDefinition of ~=
4.m∠ABD + m∠DBE + m∠EBC = m∠ABCAngle Addition Postulate
5.m∠ABD + m∠DBC = m∠ABC and m∠ABE + m∠EBC = m∠ABCAngle Addition Postulate
6.2m∠ABD = m∠ABC and 2m∠EBC = m∠ABCSubstitution (steps 3 and 5)
7.m∠ABD = m∠ABC/2 and m∠EBC = m∠ABC/2Algebra
8.m∠ABC/2 + m∠DBE + m∠ABC/2 = m∠ABCSubstitution (steps 4 and 7)
9.m∠ABC + m∠DBE = m∠ABCAlgebra
10.m∠DBE = 0Subtraction property of equality

4. The supplement of a right angle 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

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

Given: l ‌ ‌ m cut by a transversal t.

Prove: ∠1 ~= ∠3.

StatementsReasons
1. l ‌ ‌ m cut by a transversal tGiven
2. ∠1 and ∠2 are vertical anglesDefinition of vertical angles
3. ∠2 and ∠3 are corresponding anglesDefinition 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.

Given: l ‌ ‌ m cut by a transversal t.

Prove: ∠1 and ∠3 are supplementary.

StatementReasons
1. l ‌ ‌ m cut by a transversal tGiven
2. ∠1 and ∠2 are supplementary angles, and m∠1 + m∠2 = 180ºDefinition of supplementary angles
3. ∠2 and ∠3 are corresponding anglesDefinition 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 supplementaryDefinition 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.

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

Prove: l ‌ ‌ m.

StatementReasons
1.Lines l and m are cut by a transversal t, with ∠1 ~= ∠3Given
2. ∠1 and ∠2 are vertical anglesDefinition of vertical angles
3. ∠1 ~= ∠2 Theorem 8.1
4. ∠2 ~= ∠3 Transitive property of ~=.
5. ∠2 and ∠3 are corresponding anglesDefinition of corresponding angles
6.l ‌ ‌ mTheorem 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.

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

Prove: l ‌ ‌ m.

StatementReasons
1.Lines l and m are cut by a transversal t, and ∠1 are ∠3 supplementary anglesGiven
2. ∠2 and ∠1 are supplementary anglesDefinition of supplementary angles
3. ∠3 ~= ∠2 Example 2
4. ∠3 and ∠2 are corresponding anglesDefinition of corresponding angles
5.l ‌ ‌ mTheorem 10.7

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

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

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

Prove: ∠A and ∠C are complementary angles.

StatementReasons
1.ΔABC is a right triangle, and ∠B is a right angleGiven
2.m∠B = 90ºDefinition of right angle
3.m∠A + m∠B + m∠C = 180ºTheorem 11.1
4.m∠A + 90º + m∠C = 180ºSubstitution (steps 2 and 3)
5.m∠A + m∠C = 90ºAlgebra
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.

StatementReasons
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 units2

5. 30 units2

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.

StatementReasons
1. ¯AC ~= ¯CD and ∠ACB ~= ∠DCB Given
2. ¯BC ~= ¯BCReflexive property of ~=
3. ΔACB ~= ΔDCB SAS Postulate

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

StatementReasons
1. ¯CB ⊥ ¯AD and ∠ACB ~= ∠DCB Given
2. ∠ABC and ∠DBC are right anglesDefinition 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.

StatementReasons
1. ¯CB ⊥ ¯AD and ∠CAB ~= ∠CDB Given
2. ∠ABC and ∠DBC are right anglesDefinition 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.

StatementReasons
1. ¯CB ⊥ ¯AD and ¯AC ~= ¯CD Given
2. ΔABC and ΔDBC are right trianglesDefinition of right triangle
3. ¯BC ~= ¯BC Reflexive property of ~=
4. ΔACB ~= ΔDCBHL Theorem for right triangles

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

StatementReasons
1. ∠P ~= ∠R and M is the midpoint of ¯PR Given
2. ¯PM ~= ¯MR Definition of midpoint
3. ∠NMP and ∠RMQ are vertical anglesDefinition of vertical angles
4. ∠NMP ~= ∠RMQ Theorem 8.1
5. ΔPMN ~= RMQ ASA Postulate
6. ∠N ~= ∠Q CPOCTAC

### Smiliar Triangles

1. x = 11
2. x = 12
3. 40º and 140º
4. 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

1. 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.

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

Prove: E is the midpoint of ¯BC.

StatementReasons
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 anglesDefinition 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

1. AD = 63, BC = 27, RS = 45
2. ¯AX, ¯CZ, and ¯DY

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

Given: Kite ABCD.

Prove: ∠B ~= ∠D.

StatementReasons
1. ABCD is a kiteGiven
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.

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.

Given: Parallelogram ABCD.

Statement Reasons
1.Parallelogram ABCD has diagonal ¯AC. Given
2. ΔABC ~= ΔCDA Theorem 15.7

6. 144 units2

7. 180 units2

8. Kite ABCD has area 48 units2.

Parallelogram ABCD has area 150 units2.

Rectangle ABCD has area 104 units2.

Rhombus ABCD has area 35/2 units2.

### Anatomy of a Circle

1. Circumference: 20π feet, length of ˆRST = 155/18π feet
2. 9π feet2
3. 15π feet2
4. 28º

### The Unit Circle and Trigonometry

1. 3/√34 = 3√34/34
2. 1/√3 = √3/3
3. tangent ratio = √40/3, sine ratio = √40/7
4. 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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