Question

select the correct answer from each drop - down menu.
in the figure, cd
cd + de = ef +
ce = cd + de
ce = df by the
given, ab = ce and ce = df implies ab = df by the
property of equality
property of equality
property of equality

Explanation:

Step1: Identify equality setup

To relate segments, start with given: $AB = CE$, need to show $CE = DF$. First, use segment addition for $CE$ and $DF$: $CE = CD + DE$, $DF = DE + EF$.

Step2: Match segment equality

Given $CD = EF$, use Addition Property of Equality: add $DE$ to both sides:
$CD + DE = EF + DE$

Step3: Apply Segment Addition

By Segment Addition Postulate, $CD + DE = CE$ and $DE + EF = DF$.

Step4: Use Transitive Property

Substitute into the equality: $CE = DF$ by the Transitive Property of Equality (since $CD+DE=CE$, $EF+DE=DF$, and $CD+DE=EF+DE$).

Step5: Final transitive step

Given $AB = CE$ and $CE = DF$, $AB = DF$ by the Transitive Property of Equality.

Answer:

1. $CD + DE = EF + DE$: Addition Property of Equality
2. $CE = CD + DE$: Segment Addition Postulate
3. $CE = DF$: Transitive Property of Equality
4. $AB = CE$ and $CE = DF$ implies $AB = DF$: Transitive Property of Equality