Question

In the figure, CD = EF and AB = CE. Complete the statements to prove that AB = DF. CD + DE = EF + DE by the Property of Equality. CE = CD + DE and DF = EF + DE by . CE = DF by the Property of Equality. Given, AB = CE and CE = DF implies AB = DF by the Property of Equality.

Answer 1

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 a=b \ and \ b =c \ so \ a=ca=b and b=c so a=c

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

Now Given:

AB = CE

CE = DF

Now we know that if a=b \ and \ b =c \ so \ a=ca=b and b=c so a=c

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

Explanation:

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Answer 2

Answer:

Explanation:j

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