Question

objective 3: i can write proofs involving segments.

1. fill in the blanks to complete the proof.

given: b is midpoint of $overline{ac}$
$ab = x + 6$
$bc = 2x + 5$
prove: $x = 1$
word bank: (not all words listed are used)
$x=1$ $ac=bc$ subtraction property
$x+6+2x+5=0$ $ab=bc$ symmetric property
given addition property def of midpoint
(diagram: segment $ac$ with $b$ as midpoint, points $a$, $b$, $c$ collinear)

Explanation:

Step1: Use Def of midpoint

Since B is the midpoint of \( \overline{AC} \), by the Definition of Midpoint, we know that \( AB = BC \).

Step2: Substitute AB and BC expressions

We are given \( AB = x + 6 \) and \( BC = 2x + 5 \). Substituting these into \( AB = BC \), we get the equation \( x + 6 = 2x + 5 \).

Step3: Use Subtraction Property

Subtract \( x \) and \( 5 \) from both sides of the equation \( x + 6 = 2x + 5 \).
First, subtract \( x \) from both sides: \( (x + 6)-x=(2x + 5)-x \), which simplifies to \( 6=x + 5 \).
Then, subtract \( 5 \) from both sides: \( 6 - 5=(x + 5)-5 \), so \( 1=x \) or \( x = 1 \).

Answer:

The steps to prove \( x = 1 \) are:

1. By Def of midpoint, \( AB = BC \).
2. Substitute \( AB=x + 6 \) and \( BC = 2x+5 \) into \( AB = BC \) to get \( x + 6=2x + 5 \) (Given for the expressions of \( AB \) and \( BC \), and using the equality from the definition of midpoint).
3. Use the Subtraction Property on \( x + 6=2x + 5 \) to solve for \( x \), resulting in \( x = 1 \).

(If we are filling the blanks in the proof structure, the first blank (after "Given: B is midpoint of \( \overline{AC} \), \( AB=x + 6 \), \( BC = 2x + 5 \), Prove: \( x = 1 \)") would be filled with "Def of midpoint" to get \( AB = BC \), then set \( x + 6=2x + 5 \), then use "Subtraction Property" to solve \( x+6 - x-5=2x + 5-x - 5\) (or similar subtraction steps) to get \( x = 1 \).)