Question

what is the length of $overline{mn}$?
$\bigcirc$ 58 cm
$\bigcirc$ 75 cm
$\bigcirc$ 88 cm
$\bigcirc$ 116 cm
(diagram: points a, b, c, m, n with segments: $a$ to $m$ (88 cm, marked congruent), $a$ to $c$ (116 cm), $b$ to $n$ (75 cm, marked congruent), $n$ to $c$ (marked congruent), $m$ to $b$ (marked congruent))

Explanation:

Step1: Identify Midsegments

M and N are midpoints (from congruent marks), so \( \overline{MN} \) is a midsegment of \( \triangle ABC \)? Wait, no—wait, the marks: \( AM = MB \)? Wait, no, \( AM = MM' \)? Wait, looking at the diagram, \( M \) is midpoint of \( AB \)? Wait, no, \( A \) to \( M \) has one mark, \( M \) to \( B \)? Wait, no, the other side: \( BN = NC \)? Wait, no, the congruent marks on \( AB \) (wait, \( A \) to \( M \) is 88, and \( M \) to \( B \) has one mark, so \( AM = MB \)? Wait, no, the triangle: \( A \), \( C \), \( B \)? Wait, \( N \) is midpoint of \( BC \), \( M \) is midpoint of \( AB \)? Wait, no, the midsegment theorem: in a triangle, the midsegment is parallel to the third side and half its length. Wait, \( AC = 116 \), \( AB \) has \( AM = 88 \), but \( BC \) has \( BN = 75 \)? Wait, no, the congruent marks: \( M \) is midpoint of \( AB \), \( N \) is midpoint of \( BC \), so \( MN \) is midsegment, so \( MN = \frac{1}{2}AC \). Wait, \( AC = 116 \), so \( MN = \frac{116}{2} = 58 \)? No, wait, that's not matching. Wait, maybe \( MN \) is midsegment of \( \triangle ABC \), but wait, the other sides: \( AB \) has \( AM = 88 \), so \( AB = 176 \)? No, maybe I misread. Wait, the diagram: \( A \) to \( M \) is 88 (one mark), \( M \) to \( B \) has one mark (so \( AM = MB \), so \( AB = 176 \)? No, \( AC = 116 \), \( BC \) has \( BN = 75 \) (two marks), \( NC \) has two marks (so \( BN = NC \), so \( BC = 150 \))? Wait, no, the midsegment: if \( M \) is midpoint of \( AB \), \( N \) is midpoint of \( BC \), then \( MN \parallel AC \) and \( MN = \frac{1}{2}AC \). \( AC = 116 \), so \( MN = 58 \)? But the options have 58, 75, 88, 116. Wait, maybe I got the midpoints wrong. Wait, maybe \( M \) is midpoint of \( AB \), \( N \) is midpoint of \( AC \)? No, the marks on \( BC \): \( BN \) and \( NC \) have two marks, so \( BN = NC \), so \( N \) is midpoint of \( BC \). \( AM \) and \( MB \) have one mark, so \( M \) is midpoint of \( AB \). Then midsegment \( MN \) should be parallel to \( AC \) and half its length. \( AC = 116 \), so \( MN = 58 \). Wait, but let's check again. Wait, maybe the triangle is \( A \), \( C \), \( B \), with \( AC = 116 \), \( AB \) has \( AM = 88 \) (midpoint, so \( AB = 176 \)), \( BC \) has \( BN = 75 \) (midpoint, so \( BC = 150 \)). Then midsegment \( MN \) between \( AB \) and \( BC \) midpoints: \( MN = \frac{1}{2}AC = 58 \). Yes, that makes sense. So \( MN = 58 \) cm? Wait, no, wait, maybe I mixed up. Wait, the midsegment theorem: the segment connecting midpoints of two sides of a triangle is parallel to the third side and half its length. So if \( M \) is midpoint of \( AB \), \( N \) is midpoint of \( BC \), then \( MN \parallel AC \) and \( MN = \frac{1}{2}AC \). \( AC = 116 \), so \( MN = 58 \). So the answer is 58 cm.

Step2: Apply Midsegment Theorem

Using the midsegment theorem, \( MN = \frac{1}{2}AC \). Given \( AC = 116 \) cm, so \( MN = \frac{116}{2} = 58 \) cm.

Answer:

58 cm