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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