Please helppp!!!!!!


Given: AE = BD; CD = CE

Prove: AC = BC



ik it’s wrong

1) AE = BD; CD = CE

Given

2) AE = AC + CE

Subtraction Property of Equality

3) AC +CD= BC + CD

Segment Addition Postulate

4) AC +CD= BD

Symmetric Property of Equality

5) BD = AC + CD

Segment Addition Postulate

6) BD =BC+CD

Transitive Property of Equality

7) AC = BC

Substitution Property of Equality

I think this is it

To prove that AC and BC are congruent, all usable postulates and theorems must be stated, with valid reasons.

The proof that [tex]AC \cong BC[/tex] is as follows:

Given:

[tex]AE = BD, CD= CE[/tex]

I've added as an attachment, the table that needs to be completed.

[tex]1.\ AE = BD, CD= CE \to[/tex] Given

Because the given parameters are:  

[tex]AE = BD, CD= CE[/tex]

[tex]2.\ AC \cong BC \to[/tex] Definition of congruence

[tex]3.\ AE =AC + CE \to[/tex] Segment addition postulate

From the attached image, AE is a straight line, and point C is on segment AE.

[tex]4.\ BD=AC +CD \to[/tex] Substitution property of equality

Substitute BD for AE and CD for CE in (2)

[tex]5.\ AC +CD =BD \to[/tex] Symmetric property of equality

The symmetric property of equality states that:

if [tex]a = b[/tex], then [tex]b = a[/tex]

[tex]6.\ BD =BC +CD \to[/tex] Segment addition postulate

From the attached image, BD is a straight line, and point C is on segment BD.

[tex]7.\ AC+CD =BC+CD \to[/tex] Transitive property of equality

The transitive property of equality states that:

if [tex]a = b[/tex] and [tex]b = c[/tex], then [tex]a=c[/tex]

[tex]8.\ AC =BC \to[/tex] Subtraction property of equality

Subtract CD from both sides of (6)

[tex]9.\ AC \cong BC \to[/tex] Proved

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