Post on 22-Feb-2016
description
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
Page 105# 6 - 11
Homework
Page 106# 16 – 18, 21 - 22