Given: Segment AC has a length of 24 units, point \( B \) lies somewhere on segment AC and the length of segment BC is 6 units. Frove \( A B=18 \) Statement: 1. Arint \( B \) lies on segment \( A C \) 1. Given 2. \( A B+B C=A C \) 2. Postulate \( 3 . A C=24 \) 3. Given 4. \( B C=6 \) 4. Given \( 5 A B+6=24 \) 5. Substitution 6. \( A B=18 \) a. \( \square \) Select the reason that best supports Statement 6 in the given proof. Transitive Property Subtraction Property of Equality Division Property of Equality

In this scenario, the best reason that supports Statement 6 is the Subtraction Property of Equality. It allows us to subtract equal values from both sides of the equation to isolate a variable, which is exactly what happens here: \( AB + BC = AC \) leads to \( AB + 6 = 24 \). Therefore, by subtracting 6 from both sides, we determine that \( AB = 24 - 6 \), giving us \( AB = 18 \). Let’s not forget a fun fact about segments: in a straight line, any point divides the line segment into two parts, and by using these properties, we can easily figure out the lengths of those parts. So, whether you're measuring for a craft project or analyzing coordinates in geometry, these properties are your trusty sidekicks in solving linear mysteries!