Question

given that d is the midpoint of \\( \overline{ab} \\) and k is the midpoint of \\( \overline{bc} \\), which statement must be true?

\\( \bigcirc db = bk \\)
\\( \bigcirc b \\) is the midpoint of \\( \overline{ac} \\).
\\( \bigcirc d \\) bisects \\( \overline{ak} \\).
\\( \bigcirc ak + bk = ac \\)

Explanation:

Step1: Analyze the first option

Since \(D\) is the mid - point of \(AB\), \(AD = DB=\frac{1}{2}AB\). Since \(K\) is the mid - point of \(BC\), \(BK = KC=\frac{1}{2}BC\). There is no information that \(AB = BC\), so \(DB = BK\) is not necessarily true.

Step2: Analyze the second option

There is no information that \(AB=BC\). If \(AB
eq BC\), then \(B\) is not the mid - point of \(AC\).

Step3: Analyze the third option

Let \(AD = DB=x\) and \(BK = KC = y\). Then \(AK=AD + DB+BK=x + x + y=2x + y\) and \(DK=DB + BK=x + y\). Since \(AD=x
eq DK=x + y\) (unless \(y = 0\) which is not the case as \(B\) and \(C\) are distinct points), \(D\) does not bisect \(AK\).

Step4: Analyze the fourth option

By the segment addition postulate, \(AK+KC=AC\). Since \(BK = KC\) (because \(K\) is the mid - point of \(BC\)), then \(AK + BK=AC\).

Answer:

\(AK + BK = AC\)