To determine the value of \( x \) that would make the segment \( \overline{KM} \) parallel to \( \overline{JN} \), we use the converse of the side-splitter theorem. According to this theorem, if \( \frac{JK}{KL} = \frac{JM}{MN} \), then \( \overline{KM} \parallel \overline{JN} \).
Given the proportion:
[tex]\[
\frac{x-5}{x} = \frac{x-3}{x+4}
\][/tex]
we proceed as follows:
1. Set up the proportion:
By the converse of the side-splitter theorem, if:
[tex]\[
\frac{JK}{KL} = \frac{x-5}{x}
\][/tex]
and:
[tex]\[
\frac{JM}{MN} = \frac{x-3}{x+4}
\][/tex]
then \( \overline{KM} \parallel \overline{JN} \).
2. Cross-multiply:
[tex]\[
(x-5)(x+4) = x(x-3)
\][/tex]
3. Distribute:
Expand both sides:
[tex]\[
x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3)
\][/tex]
\( \Rightarrow x^2 + 4x - 5x - 20 = x^2 - 3x \)
4. Simplify:
Combine like terms:
[tex]\[
x^2 - x - 20 = x^2 - 3x
\][/tex]
5. Solve for \( x \):
Subtract \( x^2 \) from both sides to eliminate \( x^2 \):
[tex]\[
-x - 20 = -3x
\][/tex]
Add \( 3x \) to both sides:
[tex]\[
2x - 20 = 0
\][/tex]
Add 20 to both sides:
[tex]\[
2x = 20
\][/tex]
Divide by 2:
[tex]\[
x = 10
\][/tex]
Therefore, the value of \( x \) that will make \( \overline{KM} \parallel \overline{JN} \) is:
[tex]\[
x = 10
\][/tex]
[tex]\(\boxed{10}\)[/tex]
