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By Kacey Olver, Tom Jubon, and Laine Murphy
14

Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Jan 14, 2016

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Page 1: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

By Kacey Olver, Tom Jubon, and Laine Murphy

Page 2: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Section 2.4

Page 3: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Section 2.4

Page 4: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Section 2.4

Page 5: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Section 2.4

Page 6: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Statements Reasons

1. <BAE is a rt <

1.

2. 2.

3. <DEA is a rt <

3.

4. 4.

5. <CAE <CEA

5.

6. <BAC <DEC

6.

Given: <BAE is a rt < <DEA is a rt < <CAE <CEAProve: <BAC <DEC

A E

DB C

Page 7: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Given: <BAE is a rt < <DEA is a rt < <CAE <CEAProve: <BAC <DEC

A E

DB C

Statements Reasons

1. <BAE is a rt <

1. Given

2. Line FK is perpendicular to line KJ

2. If 2 lines form a rt <, then they’re perpendicular.

3. <DEA is a rt <

3. Given

4. Line JH is perpendicular to line KJ

4. Same as 2

5. <CAE <CEA

5. Given

6. <BAC <DEC

6. Compl of <‘s are

Page 8: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Given: <A is compl to <C

<DBC is compl to <C

Conclusion: __?_

A

B C

D

Page 9: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Given: <A is compl to <C

<DBC is compl to <C

Conclusion: __?_

A

B C

D

Page 10: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Statements Reasons

1. Seg KM is perp to seg MO

1.

2. <KMO is a rt < 2.

3. <RMO is compl to <KMR

3.

4. <ROM is compl to <POR

4.

5. <KMR <POR 5.

6. <ROM <RMO 6.

Given: Seg KM is perp to seg MO

Seg PO is perp to seg MO

<KMR <POR

Prove: <ROM <RMO

RK P

M O

Page 11: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Statements Reasons

1. Seg KM is perp to seg MO

1. Given

2. <KMO is a rt < 2. If segs are perp, they form rt <‘s.

3. <RMO is compl to <KMR

3. If 2 <‘s form a rt <, they are compl.

4. <ROM is compl to <POR

4. Reasons 1-3

5. <KMR <POR 5. Given

6. <ROM <RMO 6. Compl’s of <‘s are

Given: Seg KM is perp to seg MO

Seg PO is perp to seg MO

<KMR <POR

Prove: <ROM <RMO

M O

RK P

Page 12: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Given: <1 is compl to <4 <2 is compl to <3 Ray RT bisects <SRVProve: Ray TR bisects <STV

R T

S

V

1

2

3

4

Page 13: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

*Solution Provided by www.darienps.org

Statements Reasons

1. Ray RT bisects <SRV 1. Given

2. <3 <4 2. If a ray bisects an angle, then it divides the angle into 2 halves

3. <1 is compl to <4 3. Given

4. <2 is compl to <3 4. Given

5. <1 <2 5. If 2 <‘s are compl to <‘s, then they are

6. Ray TR bisects <STV 6. If a ray divides an , into 2 <‘s, then it bisects the <

Page 14: Section 2.4 Theorems By Kacey Olver, Tom Jubon, and Laine Murphy.

Rhoad, Richard, George Milauskas, and Robert Whipple. Geometry For Your Enjoyment and Challenge. MA: McDougal, Littell and Company, 1991.

Honors Geometry, Chapter 2, Packet #1, Sections 2.1-2.4

Messman, Bonita. “2.4 Congruent Supplements and Compliments.” Darien High School. 17 January 2010 < http://www.darienps.org/ teachers/bmessman/AccGeo1stQ/Ch2/Wk

4_Lesson Solutions.pdf>. Web.