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 each option

- Option 1: \(DB = BK\)
We know \(D\) is the midpoint of \(AB\), so \(AD = DB=\frac{1}{2}AB\). \(K\) is the midpoint of \(BC\), so \(BK = KC=\frac{1}{2}BC\). But we don't know if \(AB = BC\), so \(DB\) and \(BK\) may not be equal.
- Option 2: \(B\) is the midpoint of \(AC\)
For \(B\) to be the midpoint of \(AC\), \(AB\) must equal \(BC\), but there's no information given to confirm this.
- Option 3: \(D\) bisects \(\overline{AK}\)
To check if \(D\) bisects \(AK\), we need \(AD = DK\). Let's express \(DK\): \(DK=DB + BK\). Since \(AD = DB\) (as \(D\) is the midpoint of \(AB\)), we need \(DB=DB + BK\), which would imply \(BK = 0\), which is not true. Wait, no, let's re - express \(AK\). \(AK=AB + BK\), and \(AD=\frac{1}{2}AB\). \(DK=DB + BK=\frac{1}{2}AB + BK\). For \(D\) to bisect \(AK\), \(AD = DK\), so \(\frac{1}{2}AB=\frac{1}{2}AB + BK\) which is wrong. Wait, maybe I made a mistake. Let's look at the number of segments. From the diagram, \(A - D - B - K - C\). Let's assume the length of \(AD = DB=x\) and \(BK = KC = y\). Then \(AK=AD + DB+BK=2x + y\), and \(AD=x\), \(DK = DB + BK=x + y\). Wait, no, \(AK\) is from \(A\) to \(K\), so \(AK=AB + BK=2x + y\), and \(AD=x\), \(DK=DB + BK=x + y\). For \(D\) to bisect \(AK\), \(AD=\frac{1}{2}AK\), so \(x=\frac{1}{2}(2x + y)\), which simplifies to \(x=x+\frac{y}{2}\), so \(\frac{y}{2}=0\), which is not possible. Wait, maybe my initial assumption is wrong. Wait, let's count the number of equal - length segments. Since \(D\) is the midpoint of \(AB\), \(AD = DB\). Let's look at the positions: \(A\), \(D\), \(B\), \(K\), \(C\). The distance from \(A\) to \(D\) is equal to \(D\) to \(B\). Now, let's check \(AD\) and \(DK\). Wait, maybe the diagram has \(AD = DB\) and \(BK = KC\), but also, let's see the length of \(AK\). \(AK=AB + BK=2DB + BK\), and \(AD = DB\), \(DK=DB + BK\). Wait, no, if we consider the segments: \(A\) to \(D\) is 1 unit, \(D\) to \(B\) is 1 unit, \(B\) to \(K\) is 1 unit, \(K\) to \(C\) is 1 unit (maybe? But the problem doesn't state that \(AB = BC\)). Wait, maybe I misread the diagram. Wait, the problem says "which statement must be true". Let's check option 4: \(AK + BK=AC\). \(AC=AK + KC\), and since \(K\) is the midpoint of \(BC\), \(BK = KC\), so \(AC=AK + BK\). Wait, no, \(AC=AK + KC\), and \(BK = KC\), so \(AC=AK + BK\). Wait, let's re - express: \(AC=AB + BC\), \(AK=AB + BK\), and \(BC = 2BK\) (since \(K\) is the midpoint). So \(AK + BK=(AB + BK)+BK=AB + 2BK=AB + BC=AC\). Wait, but let's check option 3 again. Wait, maybe the diagram has \(AD = DB\) and \(BK = KC\), and also \(DB = BK\)? No, we don't know. Wait, let's take a step back.

Let's use the segment addition postulate.

For option 3: \(D\) is the midpoint of \(AB\), so \(AD = DB\). Let's find \(AK\): \(AK=AD + DB+BK\) (from the diagram \(A - D - B - K - C\)). So \(AK = 2DB+BK\). The midpoint of \(AK\) would be at a distance of \(\frac{AK}{2}=DB+\frac{BK}{2}\) from \(A\). But \(AD = DB\), so the distance from \(A\) to \(D\) is \(DB\), and the distance from \(D\) to the midpoint of \(AK\) is \(\frac{BK}{2}\). But if we look at the diagram, maybe \(DB = BK\)? Wait, no, the problem doesn't state that \(AB = BC\). Wait, maybe I made a mistake in option 3. Wait, let's check the options again.

Wait, the correct answer is option 3? Wait, no, let's re - examine:

Given \(D\) is the midpoint of \(AB\) (\(AD = DB\)) and \(K\) is the midpoint of \(BC\) (\(BK = KC\)).

Let's check option 3: \(D\) bisects \(\overline{AK}\). So we need \(AD = DK\).

\(DK=DB + BK\), and \(AD = DB\), so \(AD = DB=DK - BK\). For \(AD = DK\), we need \(DB=DB + BK\), which would mean \(BK = 0\), which is not possible. Wait, this is confusing. Wait, maybe the diagram is such that \(AD = DB = BK = KC\). If that's the case (even though the problem doesn't state it, but maybe from the diagram's dot positions: \(A\), \(D\), \(B\), \(K\), \(C\) with equal spacing), then \(AD = DB = BK = KC\). Then \(AK=AD + DB+BK=3\) units (if each segment is 1 unit), \(AD = 1\) unit, \(DK=DB + BK = 2\) units. No, that's not equal. Wait, maybe I misinterpret the diagram.

Wait, let's check option 4: \(AK + BK=AC\). \(AC=AK + KC\), and since \(BK = KC\) (because \(K\) is the midpoint of \(BC\)), then \(AC=AK + BK\). So this statement is always true by the segment addition postulate and the definition of midpoint (since \(BK = KC\), replacing \(KC\) with \(BK\) in \(AC = AK+KC\) gives \(AC = AK + BK\)).

Wait, but let's check the other options again:

- Option 1: \(DB = BK\): We know \(DB=\frac{1}{2}AB\) and \(BK=\frac{1}{2}BC\). We don't know if \(AB = BC\), so this is not necessarily true.
- Option 2: \(B\) is the midpoint of \(AC\): For this, \(AB = BC\), but we don't have that information.
- Option 3: \(D\) bisects \(\overline{AK}\): \(AD = DB\), \(DK=DB + BK\). For \(D\) to bisect \(AK\), \(AD = DK\), so \(DB=DB + BK\) which implies \(BK = 0\), false.
- Option 4: \(AK + BK=AC\): Since \(AC=AK + KC\) (segment addition) and \(BK = KC\) (midpoint), then \(AC=AK + BK\). This is always true.

Answer:

\(AK + BK = AC\) (the fourth option)