2.5 Proving Statements about Line Segments

Theorems are statements that can be proved

Theorem 2.1 Properties of Segment Congruence

Reflexive AB ≌ ABAll shapes are ≌ to them self

Symmetric If AB ≌ CD, then CD ≌ ABTransitive If AB ≌ CD and CD ≌ EF,

then AB ≌ EF

How to write a Proof

Proofs are formal statements with a conclusion based on given information.

One type of proof is a two column proof.

One column with statements numbered;the other column reasons that are numbered.

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.#3. EF + FG = GH + FG #3. Add. Prop.

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.#3. EF + FG = GH + FG #3. Add. Prop.#4. EG = EF + FG #4.

FH = FG + GH

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.#3. EF + FG = GH + FG #3. Add. Prop.#4. EG = EF + FG #4. Segment

Add.FH = FG + GH

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.#3. EF + FG = GH + FG #3. Add. Prop.#4. EG = EF + FG #4. Segment

Add.FH = FG + GH

#5. EG = FH #5. Subst. Prop.

Given: EF = GHProve EG ≌ FH E F G H

#1. EF = GH #1. Given#2. FG = FG #2. Reflexive

Prop.#3. EF + FG = GH + FG #3. Add. Prop.#4. EG = EF + FG #4. Segment

Add.FH = FG + GH

#5. EG = FH #5. Subst. Prop.#6. EG ≌ FH #6. Def. of ≌

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌#3. RT = RS + ST #3. Segment Add.

WY = WX + XY

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌#3. RT = RS + ST #3. Segment Add.

WY = WX + XY#4. RS + ST = WX + XY #4. Subst. Prop.

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌#3. RT = RS + ST #3. Segment Add.

WY = WX + XY#4. RS + ST = WX + XY #4. Subst. Prop.#5. ST = WX #5. Given

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌#3. RT = RS + ST #3. Segment Add.

WY = WX + XY#4. RS + ST = WX + XY #4. Subst. Prop.#5. ST = WX #5. Given#6. RS = XY #6. Subtract. Prop.

Given: RT ≌ WY; ST = WX R S T

Prove: RS ≌ XY W X Y

#1. RT ≌ WY #1. Given#2. RT = WY #2. Def. of ≌#3. RT = RS + ST #3. Segment Add.

WY = WX + XY#4. RS + ST = WX + XY #4. Subst. Prop.#5. ST = WX #5. Given#6. RS = XY #6. Subtract. Prop.#7. RS XY≌ #7. Def. of ≌

Given: x is the midpoint of MN and MX = RXProve: XN = RX

#1. x is the midpoint of MN #1. Given

Given: x is the midpoint of MN and MX = RXProve: XN = RX

#1. x is the midpoint of MN #1. Given#2. XN = MX #2. Def. of

midpoint

Given: x is the midpoint of MN and MX = RXProve: XN = RX

#1. x is the midpoint of MN #1. Given#2. XN = MX #2. Def. of

midpoint#3. MX = RX #3. Given

Given: x is the midpoint of MN and MX = RXProve: XN = RX

#1. x is the midpoint of MN #1. Given#2. XN = MX #2. Def. of

midpoint#3. MX = RX #3. Given#4. XN = RX #4. Transitive

Prop.

Something with Numbers

If AB = BC and BC = CD, then find BCA D

3X – 1 2X + 3B C

Homework

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Homework

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