Question

question 6 of 8
select the correct answer from each drop - down menu.
given: \\( \overline { a c } \cong \overline { x z } , \overline { b c } \cong \overline { x y } \\)
prove: \\( \overline { a b } \cong \overline { y z } \\)
complete the proof.
suppose \\( \overline { a b } \
ot \cong \overline { y z } \\). by the definition of congruence, \\( a b \
eq y z \\). by the addition property of equality, \\( a b + b c \
eq y z + b c \\). using the given, \\( b c = x y \\) because of the definition of congruence. applying the substitution property of equality, \\( a b + b c \
eq y z + x y \\). applying \\( a b + b c \
eq y z + x y \\),
by the definition of congruence. thus, \\( \overline { a b } \cong \overline { y z } \\).

Explanation:

Step1: Understand 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 \(AC = AB+BC\). Similarly, if \(X\), \(Y\), and \(Z\) are collinear points with \(Y\) between \(X\) and \(Z\), then \(XZ=XY + YZ\).

Step2: Analyze the given congruences and the inequality

We are given \(\overline{AC}\cong\overline{XZ}\) (so \(AC = XZ\) by the definition of congruent segments) and \(\overline{BC}\cong\overline{XY}\) (so \(BC=XY\) by the definition of congruent segments). We have the inequality \(AB + BC
eq YZ+BC\) (from \(AB
eq YZ\) using the addition property of equality), and then \(AB + BC
eq YZ + XY\) (by substitution since \(BC = XY\)). But if we apply the segment addition postulate: \(AC=AB + BC\) and \(XZ=XY+YZ\). Since \(AC = XZ\) (given \(\overline{AC}\cong\overline{XZ}\)), if \(AB + BC
eq YZ+XY\), we get a contradiction. The key step is using the segment addition postulate which relates the lengths of the composed segments.

Answer:

segment addition