Question

proof write a two - column proof to prove each geometric relationship.

1. if \\( \overline { vz } \cong \overline { vy } \\) and \\( \overline { wy } \cong \overline { xz } \\).

then \\( \overline { vw } \cong \overline { vx } \\).

1. if \\( e \\) is the midpoint of \\( \overline { df } \\) and

\\( \overline { cd } \cong \overline { fg } \\), then \\( \overline { ce } \cong \overline { eg } \\).
example 3

1. family maria is 11 inches shorter than her sister clara. luna is 11 inches shorter than her

brother chad. if maria is shorter than luna, how do the heights of clara and chad compare?
what else can be concluded if maria and luna are the same height?

1. lumber byron works in a lumberyard. his boss just cut a dozen planks and asked byron to

double check that they are all the same length. the planks were numbered 1 through 12. byron
took out plank number 1 and checked that the other planks are all the same length as plank 1.
he concluded that they must all be the same length. explain how you know that plank 7 and
plank 10 are the same length even though they were never directly compared to each other.

1. neighborhoods karla, lola, and mandy live in three houses that are on the same line. lola

lives between karla and mandy. karla and mandy live a mile apart. is it possible for lolas house
to be a mile from both karlas and mandys houses?
mixed exercises

1. proof five lights, \\( a, b, c, d \\), and \\( e \\), are aligned in a row. the middle light is the midpoint of the

segment between the second and fourth lights and also the midpoint of the segment between
the first and last lights.
a. draw a figure to illustrate the situation.
b. complete this proof.
given: \\( c \\) is the midpoint of \\( \overline { bd } \\) and \\( \overline { ae } \\).
prove: \\( a b = d e \\)
\

$$\begin{tabular}{|c|c|}hline statement & reason \\hline 1. \\( c \\) is the midpoint of \\( \\overline { bd } \\) and \\( \\overline { ae } \\). & 1. given \\hline 2. \\( b c = c d \\) and \\( \\) ? & 2. \\( \\) ? \\hline 3. \\( a c = a b + b c , c e = c d + d e \\) & 3. \\( \\) ? \\hline 4. \\( a c - b c = a b \\) & 4. \\( \\) ? \\hline 5. \\( \\) ? & 5. substitution property \\hline 6. \\( c e - c d = d e \\) & 6. \\( \\) ? \\hline 7. \\( a b = c e - c d \\) & 7. symmetric property of equality \\hline 8. \\( \\) ? & 8. \\( \\) ? \\hline\\end{tabular}$$

date
period

Explanation:

Step 1: Use the mid - point definition

Since \(C\) is the mid - point of \(\overline{BD}\) and \(\overline{AE}\), by the definition of a mid - point, if \(C\) is the mid - point of a segment \(\overline{XY}\), then \(XC = CY\). So, \(BC = CD\) and \(AC=CE\).

Step 2: Use the segment addition postulate

The segment addition postulate states that if \(A\), \(B\), and \(C\) are collinear points with \(B\) between \(A\) and \(C\), then \(AC=AB + BC\) and \(CE=CD + DE\).

Step 3: Solve for \(AB\) from \(AC = AB+BC\)

Subtract \(BC\) from both sides of the equation \(AC = AB + BC\). Using the subtraction property of equality (\(a=b\) implies \(a - c=b - c\)), we get \(AC - BC=AB\).

Step 4: Substitute \(AC\) with \(CE\)

Since \(AC = CE\) (from step 1), by the substitution property, we can write \(CE - BC=AB\).

Step 5: Solve for \(DE\) from \(CE = CD + DE\)

Subtract \(CD\) from both sides of the equation \(CE = CD + DE\). Using the subtraction property of equality, we get \(CE - CD=DE\).

Step 6: Conclude \(AB = DE\)

Since \(AB=CE - BC\) and \(DE = CE - CD\) and \(BC = CD\) (from step 1), by the substitution property (if \(a=b\) and \(c = d\) and \(a - c\) and \(b - d\) are considered, here \(a = CE\), \(b = CE\), \(c = BC\), \(d = CD\)), we have \(AB=DE\).

Answer:

1. Given
2. \(AC = CE\); Definition of mid - point
3. Segment Addition Postulate
4. Subtraction Property of Equality
5. \(CE - BC=AB\)
6. Subtraction Property of Equality
7. Symmetric Property of Equality
8. \(AB = DE\); Substitution Property (because \(BC = CD\) and from steps 5 and 6)