Question

given (overline{ed}congoverline{db}), which statements about the figure are true? check all that apply.
(squareoverline{eb}) is bisected by (overline{df}).
(square) a is the midpoint of (overline{fc}).
(squareoverline{fc}) bisects (overline{db}).
(squareoverline{eb}) is a segment bisector.
(square fa=\frac{1}{2}fc).
(squareoverline{da}congoverline{ab}).

Explanation:

Step1: Analyze the mid - point and segment bisector concepts

Since \( \overline{ED}\cong\overline{DB}\), point \(D\) is on \( \overline{EB}\). If \( \overline{FC}\) has a point \(A\) such that \(CA = AB\) (marked with the same tick marks), then \(A\) is the mid - point of \( \overline{CB}\). Also, if \(A\) is the mid - point of \( \overline{FC}\) (assuming the tick marks on \( \overline{FC}\) imply \(FA=AC\)), and \(FA=\frac{1}{2}FC\) (because if \(FA = AC\), then \(FC=FA + AC=2FA\)).

Step2: Check each option

• For \( \overline{DA}\cong\overline{AB}\): There is no information or geometric property (like congruent triangles etc.) given in the problem to support this congruence.
• For \(FA=\frac{1}{2}FC\): If \(A\) is the mid - point of \( \overline{FC}\) (from the segment bisector markings on \( \overline{FC}\)), then by the definition of a mid - point \(FA = AC\) and \(FC=FA + AC\), so \(FA=\frac{1}{2}FC\).
• For \( \overline{EB}\) is a segment bisector: A segment bisector is a line, ray, or segment that divides a segment into two equal parts. There is no indication that \( \overline{EB}\) divides another segment into two equal parts.
• For \( \overline{FC}\) bisects \( \overline{DB}\): There is no information about the relationship between \( \overline{FC}\) and \( \overline{DB}\) in terms of bisecting (no equal sub - segments of \( \overline{DB}\) related to \( \overline{FC}\) are shown).
• For \(A\) is the mid - point of \( \overline{FC}\): The tick marks on \( \overline{FC}\) (assuming they represent equal lengths) imply that \(FA = AC\), so by the definition of a mid - point, \(A\) is the mid - point of \( \overline{FC}\).
• For \( \overline{EB}\) is bisected by \( \overline{DF}\): Just because \(ED\cong DB\) (given \( \overline{ED}\cong\overline{DB}\)), but there is no information that \( \overline{DF}\) is involved in bisecting \( \overline{EB}\) (no intersection point of \( \overline{DF}\) and \( \overline{EB}\) with equal - length sub - segments of \( \overline{EB}\) from the intersection is shown).

Answer:

\(FA=\frac{1}{2}FC\), \(A\) is the mid - point of \( \overline{FC}\)