Question

given that d is the midpoint of \\(\overline{ab}\\) and k is the midpoint of \\(\overline{bc}\\), which statement must be true?\
\
a d b k c\
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\\(\bigcirc\\) \\(db = bk\\)\
\\(\bigcirc\\) b is the midpoint of \\(\overline{ac}\\).\
\\(\bigcirc\\) d bisects \\(\overline{ak}\\).\
\\(\bigcirc\\) \\(ak + bk = ac\\)

Explanation:

Step1: Analyze midpoint definitions

Since D is the midpoint of \( \overline{AB} \), we have \( AD = DB=\frac{1}{2}AB \). Since K is the midpoint of \( \overline{BC} \), we have \( BK = KC=\frac{1}{2}BC \). But we don't know if \( AB = BC \), so \( DB \) and \( BK \) may not be equal.

Step2: Check if B is midpoint of \( \overline{AC} \)

From the diagram, \( AB \) and \( BC \) are segments with B between A and C, but there's no indication \( AB = BC \), so B is not necessarily the midpoint of \( \overline{AC} \).

Step3: Check if D bisects \( \overline{AK} \)

Let's denote \( AD = DB = x \) and \( BK = KC = y \). Then \( AK=AD + DB+BK=x + x + y = 2x + y \), and the midpoint of \( AK \) would be at \( x+\frac{y}{2} \) from A. But \( AD = x \) and \( DB = x \), so D is at \( x \) from A, which is not the midpoint unless \( y = 0 \), which it's not. Wait, maybe I made a mistake. Wait, the points are A---D---B---K---C. So \( AK = AB + BK \), and \( AB = 2DB \), \( BK \) is just \( BK \). Wait, no, let's take lengths. Let \( AD = DB = m \), so \( AB = 2m \). Let \( BK = KC = n \), so \( BC = 2n \). Then \( AK=AB + BK=2m + n \), and \( AD + DK= m + (DB + BK)=m+(m + n)=2m + n \). Wait, no, \( DK = DB + BK=m + n \), so \( AD = m \), \( DK = m + n \), so \( AD eq DK \) unless \( n = 0 \). Wait, maybe the correct approach is to check the last option.

Step4: Check \( AK + BK = AC \)

We know that \( AC=AB + BC \), and \( AB = AD + DB \), \( BC = BK + KC \). But \( AK=AB + BK \) (since A to K is A to B to K). So \( AK + BK=(AB + BK)+BK=AB + 2BK \), which is not equal to \( AC=AB + BC=AB + 2BK \) (since \( BC = 2BK \) as K is midpoint). Wait, \( AC = AB + BC=AB + 2BK \), and \( AK = AB + BK \), so \( AK + BK=(AB + BK)+BK=AB + 2BK = AC \). Wait, that works. Wait, let's re-express:

\( AC = AB + BC \)

Since K is midpoint of \( BC \), \( BC = 2BK \), so \( AC = AB + 2BK \)

\( AK = AB + BK \) (because A to K is A to B to K)

So \( AK + BK=(AB + BK)+BK=AB + 2BK = AC \)

Now check the third option: D bisects \( \overline{AK} \). \( AK = AD + DB + BK \), and \( AD = DB \) (since D is midpoint of AB), so \( AK = 2DB + BK \). The midpoint of \( AK \) would be at \( \frac{AK}{2}=DB+\frac{BK}{2} \). The length from A to D is \( DB \), and from D to midpoint is \( \frac{BK}{2} \), and from midpoint to K is \( DB+\frac{BK}{2} \)? Wait, no, maybe I messed up the third option. Wait, let's check the first option again. \( DB = BK \): \( DB=\frac{AB}{2} \), \( BK=\frac{BC}{2} \), and we don't know if \( AB = BC \), so \( DB \) and \( BK \) may not be equal. Second option: B is midpoint of AC? \( AC = AB + BC \), B is midpoint only if \( AB = BC \), which we don't know. Third option: D bisects AK? \( AK = AD + DB + BK \), \( AD = DB \), so \( AK = 2DB + BK \). The midpoint of AK would be at \( DB+\frac{BK}{2} \) from A. But \( AD = DB \), so D is at \( DB \) from A, and the midpoint is at \( DB+\frac{BK}{2} \), so D is not the midpoint unless \( BK = 0 \). Fourth option: \( AK + BK = AC \). \( AK = AB + BK \), so \( AK + BK = AB + BK + BK = AB + 2BK \). But \( AC = AB + BC \), and since K is midpoint of BC, \( BC = 2BK \), so \( AC = AB + 2BK \). Therefore, \( AK + BK = AC \) is true. Wait, but let's check with the diagram. A---D---B---K---C. So \( AK \) is from A to K, \( BK \) is from B to K, so \( AK + BK \) would be A to K to B to K? No, wait, \( AK \) is A to K, and \( BK \) is B to K, so \( AK + BK = (A to K)+(B to K)= (A to B to K)+(B to K)=A to B to K to B to K \), which is not correct. Wait, no, length is a scalar. So \( AK \) is the length from A to K, \( BK \) is length from B to K. So \( AK = AB + BK \), so \( AK + BK = AB + BK + BK = AB + 2BK \). \( AC = AB + BC = AB + 2BK \) (since K is midpoint, \( BC = 2BK \)). So yes, \( AK + BK = AC \) is true. Wait, but earlier I thought the third option was wrong, but let's re-examine the third option: "D bisects \( \overline{AK} \)". \( AK \) is from A to K. The midpoint of \( AK \) would be a point M such that \( AM = MK \). Let's take coordinates. Let A be at 0, D at m, B at 2m, K at 2m + n, C at 2m + 2n. Then \( AK \) is from 0 to \( 2m + n \), midpoint at \( m+\frac{n}{2} \). D is at m, so \( AM = m+\frac{n}{2} \), \( DM=\frac{n}{2} \), \( MK=(2m + n)-(m+\frac{n}{2})=m+\frac{n}{2} \). Wait, \( AM = MK \), so D is at m, and midpoint is at \( m+\frac{n}{2} \), so D is not the midpoint unless \( n = 0 \). So the third option is wrong. The fourth option: \( AK + BK = AC \). \( AK = 2m + n \), \( BK = n \), so \( AK + BK = 2m + n + n = 2m + 2n \). \( AC = 2m + 2n \) (since \( AC = AB + BC = 2m + 2n \)). So yes, \( AK + BK = AC \) is true. Wait, but let's check the first option: \( DB = BK \). \( DB = m \), \( BK = n \), we don't know if m = n, so not necessarily true. Second option: B is midpoint of AC? \( AC = 2m + 2n \), midpoint would be at \( m + n \), but B is at 2m, so 2m = m + n implies m = n, which we don't know. So the correct answer is the last option: \( AK + BK = AC \). Wait, but let's check again. \( AC = AB + BC \), \( AB = 2DB \), \( BC = 2BK \), so \( AC = 2DB + 2BK \). \( AK = AB + BK = 2DB + BK \), so \( AK + BK = 2DB + BK + BK = 2DB + 2BK = AC \). Yes, that's correct.

Answer:

\( AK + BK = AC \) (the last option, but since the options are not labeled with letters here, but in the original problem, the last option is "AK + BK = AC")