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olasank [31]
3 years ago
7

Which statement shows the Transitive Property of Equality? a. If AB = CD and AB =EF, then CD is not Equal to EF. b. If AB + BC =

DE + BC, then AB = DE. c. IF AB = CD, and CD = DE, then AB = DE d. If AB = CD, then CD = AB.
Mathematics
2 answers:
steposvetlana [31]3 years ago
8 0
The answer is c
AB = CD, CD = DE the AB = DE
Alex777 [14]3 years ago
7 0

Answer:

c. If AB=CD and CD= DE ,then AB=DE.

Step-by-step explanation:

Statement of Transitive property of equality : It states that If two numbers are equal to third number then they are also equal to each other.

In short

 If a=b  and b= c

then a= c

a. AB= CD and AB=EF

Then CD=EF

But we are given that CD is not equal to EF. Therefore, it is false.

b.If AB+BC=DE+BC

Then , AB=DE , It is not a statement of transitive property of equality . When both sides added equal part of segment then the segments are equal.Therefore, it is not a transitive property. Hence,it is false.

c.If AB=CD and CD= DE

Then AB=DE

It follows the statement of transitive property . Therefore, it is true.

d. If AB=CD then CD=AB

It is a commutative property of equality . Therefore, it is not a statement of transitive property of equality.Hence, it is false.

Answer : c. If AB=CD and CD= DE ,then AB=DE.

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