# Theorems are statements that can be proved Theorem 2.1 Properties of Segment Congruence ReflexiveAB ≌ AB All shapes are ≌ to them self Symmetric If AB.

## Presentation on theme: "Theorems are statements that can be proved Theorem 2.1 Properties of Segment Congruence ReflexiveAB ≌ AB All shapes are ≌ to them self Symmetric If AB."— Presentation transcript:

Theorems are statements that can be proved Theorem 2.1 Properties of Segment Congruence ReflexiveAB ≌ AB All shapes are ≌ to them self Symmetric If AB ≌ CD, then CD ≌ AB Transitive 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 = GH Prove EG ≌ FH EFGH #1.EF = GH#1.Given

Given: EF = GH Prove EG ≌ FH EFGH #1.EF = GH#1.Given #2.FG = FG#2. Reflexive Prop.

Given: EF = GH Prove EG ≌ FH EFGH #1.EF = GH#1.Given #2.FG = FG#2. Reflexive Prop. #3.EF + FG = GH + FG#3. Add. Prop.

Given: EF = GH Prove EG ≌ FH EFGH #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 = GH Prove EG ≌ FH EFGH #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 = GH Prove EG ≌ FH EFGH #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 = GH Prove EG ≌ FH EFGH #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 RS T Prove: RS ≌ XY W X Y #1.RT ≌ WY#1.Given

Given: RT ≌ WY; ST = WX RS T Prove: RS ≌ XY W X Y #1.RT ≌ WY#1.Given #2.RT = WY#2.Def. of ≌

Given: RT ≌ WY; ST = WX RS 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 RS 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 RS 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 RS 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 RS 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 = RX Prove:XN = RX #1.x is the midpoint of MN#1.Given

Given:x is the midpoint of MN and MX = RX Prove: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 = RX Prove: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 = RX Prove: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 BC A D 3X – 12X + 3 B C

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