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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, or transitive)
Property of Equality.

CE=CD + DE and DF = EF + DE by (addition,subtraction, segment addition, or transitive)

CE = DF by the (addition,subtraction, segment addition, or transitive) Property of Equality.

Given, AB = CE and CE = DF implies AB = DF by the (addition,subtraction, segment addition, or transitive) Property of Equality.

In the figure CDEF and AB CE Complete the statements to prove that AB DF CD DE EF DE by the additionsubtractionsubstitution or transitive Property of Equality class=

Respuesta :

Answer:

Hence Proved AB = DF

Step-by-step explanation:

In the Figure:

Given;

CD = EF

AB = CE

We need to prove AB = DF

Solution:

CD = EF    ⇒ (Given)

Now Adding both side by DE we get;

CD + DE = EF + DE   ⇒by the (Addition) Property of Equality.

CE=CD + DE and DF = EF + DE ⇒(Segment Addition)

Now we know that if [tex]a=b \ and \ b =c \ so \ a=c[/tex]

CE = DF           ⇒by the (transitive) Property of Equality.

Now Given:

AB = CE

CE = DF

Now we know that if [tex]a=b \ and \ b =c \ so \ a=c[/tex]

AB  = DF    ⇒by the (Transitive) Property of Equality)

abidemiokin

The correct options to fill in the gaps are:

  • Addition postulate
  • Segment Addition
  • Transitive Property of Equality
  • Transitive Property of Equality

From the diagram given, we have that;

  • CD = EF
  • AB = CE

We are to show that the segment AB is congruent to DF

Also from the diagram

  • CD + DE = EF + DE according to the Addition postulate of Equality

  • CE = CD + DE and DF = DE + EF according to the Segment Addition

  • Since CD = EF, hence DF = DE + CE, this means

  • CD = DF by the Transitive Property of Equality

Similarly, given that:

AB = CE and CE = DF implies AB = DF by the Transitive Property of Equality.

Learn more here: https://brainly.com/question/13044549

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