Question

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

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

Explanation:

Step1: Analyze each option

• Option 1: \(DB = BK\)

There is no information given that \(AB = BC\), so we cannot conclude \(DB = BK\) (since \(D\) is the mid - point of \(AB\) (\(AD = DB=\frac{1}{2}AB\)) and \(K\) is the mid - point of \(BC\) (\(BK = KC=\frac{1}{2}BC\))).

• Option 2: \(B\) is the midpoint of \(\overline{AC}\)

There is no information given that \(AB = BC\), so we cannot say \(B\) is the mid - point of \(AC\).

• Option 3: \(D\) bisects \(\overline{AK}\)

Let \(AD = DB=x\) and \(BK = KC = y\). Then \(AK=AB + BK=2x + y\) and \(AD=x\). Since \(x
eq\frac{2x + y}{2}\) (in general, unless \(y = 0\), but \(y>0\) as \(K\) is a distinct point from \(B\)), \(D\) does not bisect \(AK\).

• Option 4: \(AK+BK = AC\)

By the segment addition postulate, \(AK=AB + BK\) and \(AC=AB + BC\). Since \(K\) is the mid - point of \(BC\), \(BC = 2BK\). Then \(AK+BK=(AB + BK)+BK=AB + 2BK=AB + BC=AC\).

Answer:

\(AK + BK=AC\)