WARM-UP

Rewrite each of the following statements in “If-then” form as the conditional, and converse. then write a biconditional and determine if it is . is true or false. 1. 𝑥 2 = 64; 𝑥 = −8 2. Vietnamese New Years is on January 3. 3. AB+BC=AC, B is between AC

CAHSEE prep

GEOMETRY GAME PLAN Date Section / Topic Lesson Goal 9/30/13 Monday Notes: 2.5 Proving Statements about segments Students will be able to write proofs with reasons about congruent segments .

Geometry California Standard 2.0

Students write geometric proofs.

Homework Announcements P. 104-107 (#1-11, 16, 31-34) Math tutoring is available every Mon-Thurs in Room 307, 3-4PM! Test next Tuesday.

PROVING STATEMENTS ABOUT SEGMENTS

What is the difference between congruence and equality?

Building a Proof

 When writing a proof, you can only use facts that have previously been proved (theorems), facts that are assumed true without proof (postulates), and definitions .

 Proofs can be written in paragraph form or in a

two-column form.

 We will use two-column form most often.

Two-Column Proofs: Key Elements

① ② ③ ④ ⑤

Given

: state the “given” facts

Diagram

: a figure that shows what is given

Prove

: a statement of what you have to prove

Statements

: (Left Column) numbered logical statements that lead to your conclusion

Reasons

: (Right Column) numbered reasons that justify your statements (definitions, postulates, properties of algebra/congruence, previously proven theorems)

Properties of Equality (review)



These two properties are often interchangeable:

 Substitution Prop. of Equality: If a=b, then we can substitute (plug in) a for b, or b for a.

 If x+a=c AND a=b, then x+b=c (Plug it in, plug it in!)  Transitive Prop. of Equality: If a=b & b=c, then a=c.

 If Mr. Madden is the same height as Simon, and Simon is the same height a Bradley, then Mr. Madden and Bradley are the same height. =)

Properties of Equality (review cont.)

 Mirror, mirror, on the wall…  Reflexive Prop. of Equality: a=a  Anything is equal to itself (Think about your reflection in the mirror!)  Symmetric Property of Equality: If a=b, then b=a.

 Think about something symmetrical… if you flip it, it still looks the same. You can always flip an equation, Left  to  Right. (Flip it good!)

Properties of Congruence



These 3 properties work for congruence also:

 Reflexive: For any segment AB, AB ≅ AB.

 Symmetric: If AB ≅ CD, then CD ≅ AB.

 Transitive: If AB ≅ CD & CD ≅ EF, then AB ≅ EF.

Given: AB = BC, C is the midpoint of BD Prove: AB = CD

Statement

1. AB = BC, C is the midpoint of BD

2.

BC = CD

3.

AB = CD

Reason

1.

Given

2.

Def. Midpoint

3.

Substitution

Given: AB=CD Prove: AC=BD

Statement

1. AB=CD

2.

AB+BC=AC

3.

BC+CD=BD

4. 5.

BC+ AB =BD AC=BD

Reason

1. Given 2. Segment Add. Post. 3. Segment Add. Post.

4. Substitution 5. Substitution

Given: AC=BD Prove: AB=CD

Statement

1. AC=BD

2.

AB+BC=AC

3.

BC+CD=BD

4. 5.

AB+BC = BC+CD AB=CD

Reason

1. Given 2. Segment Add. Post.

3. Segment Add. Post.

4. Substitution 5. Subtr. Prop of =

Given: QR = RS Prove: QS = 2 RS Statement

1.

2. 3. 4.

Reason

1. 2. 3. 4.

Given:

LE = RM, EG = AR

Prove: LG = MA Statement

1. LE = RM and EG = AR

Reason

2.

AR+RM=AM

3.

LE+EG=LG

3.

Segment addition

4.

RM+AR=LG Substitution

5.

LG = MA

5.

Substitution

Example 5: Using Segment Relationships

• • • In the diagram, Q is the midpoint of PR. Show that PQ and QR are equal to ½ PR.

GIVEN: Q is the midpoint of PR.

PROVE: PQ = ½ PR and QR = ½ PR.

R Q P

1.

2.

3.

4.

5.

6.

7.

Statements: Reasons: Q is the midpoint of PR.

PQ = QR PQ + QR = PR PQ + PQ = PR 2 ∙ PQ = PR PQ = ½ PR QR = ½ PR 1.

2.

3.

4.

5.

6.

7.

Given Definition of a midpoint Segment Addition Postulate Substitution Property Distributive property Division property Substitution

(over Lesson 2-2) Write using two column proofs!

Slide 1 of 1

(over Lesson 2-2) Slide 1 of 1

Example 2: Using Congruence

• • • Use the diagram and the given information to complete the missing steps and reasons in the proof.

GIVEN: LK = 5, JK = 5, JK ≅ JL PROVE: LK ≅ JL K J L

Statements: Reasons: 1.

2.

3.

4.

5.

6.

________________ ________________ LK = JK LK ≅ JK JK ≅ JL ________________ 1.

2.

3.

4.

5.

6.

Given Given Substitution _________________ Given Substitution

Statements: Reasons: 1.

2.

3.

4.

5.

6.

LK = 5 JK = 5 LK = JK LK ≅ JK JK ≅ JL LK ≅ JL 1.

2.

3.

4.

5.

6.

Given Given Substitution Def. Congruent seg.

Given Substitution

AB = BC DE = EF

Def. of congruent segments Substitution (or Transitive) Substitution (or Transitive) Def. of congruent segments