In the figure, cd = ef and ab = ce. complete the statements to prove that ab = df.

cd + de = ef + de by the (addition, subtraction, substitution, transitive) property of equality.

ce = cd + de and df = ef + de by (addition, subtraction, segment addition, transitive).

ce = df by the (addition, subtraction, substitution, transitive)property of equality.

given, ab = ce and ce = df implies ab = df by the (addition, subtraction, substitution, transitive)property of equality.

1. Addition Property of Equality

2. Segment addition

3. Substitution Property of Equality.

4. Transitive Property of Equality.

Step-by-step explanation:

Here, given : CD = EF and AB = CE

To Show:  AB = DF

Now, as given in the steps:

1. CD + DE = EF + DE by the (addition) Property of Equality.

As we have added the EQUAL QUANTITY on both sides of the equality.

2.CE = CD + DE and DF = EF + DE by (segment addition).

As here CE and DF are line segments. And the length of  a

Line Segment = Sum of all its parts in which it is divided.

3.CE = DF by the (Addition, subtraction, substitution, transitive)Property of Equality.

Here, as  we know CE  = CD  + DE

but CD  = EF , so SUBSTITUTE EF in place of CD

⇒ CE = EF + ED  = FD (by substitution) Property of Equality.

4.Given, AB = CE and CE = DF implies AB = DF by the (transitive)Property of Equality.

As, x = y , y = z ⇒ x = z is the TRANSITIVE PROPERTY

answers is c

Step-by-step explanation:

CD=EF                  > given

CD+ DE= EF+ DE     > add DE to both side

CE = CD+DE      

DF= EF+ DE     

CE=DF     > transitive property of equality, since CD+DE = EF+ DE 

Since CE=DF and AB=CE (given), then AB=DF 

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

could you provide figure.