Question

proof a city planner is designing a new park. the park has two straight paths, (overline{ab}) and (overline{cd}), which are the same length. a monument, ( m ), is located at the midpoint of both paths.
image of segments ( a-m-b ) and ( c-m-d ) intersecting at ( m )
( \boldsymbol{a} ). the city planner thinks that the length of (overline{am}) will be the same as the length of (overline{cm}). explain why this makes sense.
both segments are half the length of (\boldsymbol{\text{select choice}}), so the lengths of the shorter segments must be (\boldsymbol{\text{select choice}}).

Explanation:

Step1: Recall Midpoint Definition

A midpoint of a segment divides it into two equal - length sub - segments. For segment \(\overline{AB}\), since \(M\) is the midpoint, \(AM=\frac{1}{2}AB\). For segment \(\overline{CD}\), since \(M\) is the midpoint, \(CM = \frac{1}{2}CD\).

Step2: Use Given Information

We are given that \(AB = CD\) (the two paths are the same length).

Step3: Compare \(AM\) and \(CM\)

Since \(AM=\frac{1}{2}AB\) and \(CM=\frac{1}{2}CD\), and \(AB = CD\), then by the substitution property (substituting \(AB\) with \(CD\) in the equation for \(AM\)), we have \(AM = CM\).

For the first "Select Choice" box: The two segments \(AM\) and \(CM\) are half the length of \(\overline{AB}\) (or \(\overline{CD}\), since \(AB = CD\)). So we can choose \(\overline{AB}\) (or \(\overline{CD}\)).
For the second "Select Choice" box: Since \(AM=\frac{1}{2}AB\) and \(CM=\frac{1}{2}CD\) and \(AB = CD\), the lengths of the shorter segments (\(AM\) and \(CM\)) must be equal.

Answer:

First "Select Choice" box: \(\overline{AB}\) (or \(\overline{CD}\))
Second "Select Choice" box: equal