given (overline{ed}congoverline{cd}). 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}).

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Accepted Answer

# Brief Explanations:
1. **$\boldsymbol{\overline{EB}}$ is bisected by $\boldsymbol{\overline{DF}}$**: We know $\overline{ED}\cong\overline{DB}$ (given $\overline{ED}\cong\overline{DB}$? Wait, no, the given is $\overline{ED}\cong\overline{DB}$? Wait, the problem says $\overline{ED}\cong\overline{DB}$? Wait, the problem states "Given $\overline{ED}\cong\overline{DB}$"? Wait, no, looking at the figure, the markings: $\overline{DF}$ and $\overline{FC}$ have a mark on $\overline{FA}$ and $\overline{AC}$? Wait, no, let's re-express. The given is $\overline{ED}\cong\overline{DB}$? Wait, the problem says "Given $\overline{ED}\cong\overline{DB}$" (wait, the original problem: "Given $\overline{ED}\cong\overline{DB}$"? Wait, the user's image: "Given $\overline{ED}\cong\overline{DB}$" – no, the user's text: "Given $\overline{ED}\cong\overline{DB}$" – wait, the options: Let's analyze each:

- **$\overline{EB}$ is bisected by $\overline{DF}$**: For a segment to be bisected, the bisector must intersect it at its midpoint. Since $\overline{ED}\cong\overline{DB}$, $D$ is the midpoint of $\overline{EB}$, so if $\overline{DF}$ passes through $D$, then $\overline{DF}$ bisects $\overline{EB}$. Wait, but the figure: $D$ is on $\overline{EB}$, and $\overline{DF}$ connects $D$ to $F$. So yes, $D$ is the midpoint (because $\overline{ED}\cong\overline{DB}$), so $\overline{DF}$ bisects $\overline{EB}$. Wait, no: the statement is "$\overline{EB}$ is bisected by $\overline{DF}$" – a segment is bisected by another segment if the other segment intersects it at its midpoint. Since $D$ is the midpoint of $\overline{EB}$ (because $\overline{ED}\cong\overline{DB}$), and $\overline{DF}$ passes through $D$, so $\overline{DF}$ bisects $\overline{EB}$. So this is true? Wait, maybe I misread. Wait, the given is $\overline{ED}\cong\overline{DB}$, so $D$ is midpoint of $\overline{EB}$. Then $\overline{DF}$ goes through $D$, so $\overline{EB}$ is bisected by $\overline{DF}$? Wait, no: the bisector of $\overline{EB}$ would be a segment that intersects $\overline{EB}$ at its midpoint. So $\overline{DF}$ intersects $\overline{EB}$ at $D$, which is the midpoint (since $\overline{ED}\cong\overline{DB}$), so yes, $\overline{EB}$ is bisected by $\overline{DF}$? Wait, maybe not. Wait, the first option: "$\overline{EB}$ is bisected by $\overline{DF}$" – let's check.

- **A is the midpoint of $\overline{FC}$**: The markings on $\overline{FC}$: $\overline{FA}$ and $\overline{AC}$ have the same tick mark, so $\overline{FA}\cong\overline{AC}$, so $A$ is the midpoint of $\overline{FC}$. Wait, but the option says "A is the midpoint of $\overline{FC}$" – yes, because $\overline{FA}\cong\overline{AC}$, so $A$ is midpoint. Wait, but let's check other options.

- **$\overline{FC}$ bisects $\overline{EB}$**: $\overline{FC}$ intersects $\overline{EB}$ at $D$? No, $\overline{FC}$ intersects $\overline{EB}$ at $A$? Wait, no, the figure: $\overline{EB}$ is a horizontal line (from $E$ to $B$), $\overline{FC}$ is a line from $F$ to $C$, intersecting $\overline{EB}$ at $A$. Wait, $D$ is on $\overline{EB}$, left of $A$. Wait, I think I misread the given: the problem says "Given $\overline{ED}\cong\overline{DB}$" – so $D$ is midpoint of $\overline{EB}$ (since $ED = DB$). Then $\overline{FC}$ intersects $\overline{EB}$ at $A$. For $\overline{FC}$ to bisect $\overline{EB}$, $A$ would have to be midpoint of $\overline{EB}$, but $D$ is midpoint (since $ED = DB$), so $A
eq D$ (from figure), so $\overline{FC}$ does not bisect $\overline{EB}$. So that option is false.

- **$\overline{EB}$ is a segment bisector**: A segment bisector is a segment that bisects another segment. $\overline{EB}$: does it bisect any segment? Wait, $\overline{EB}$ has midpoint $D$. If $\overline{EB}$ bisects another segment, say $\overline{FC}$, but $\overline{EB}$ intersects $\overline{FC}$ at $A$, and $A$ is midpoint of $\overline{FC}$ (since $\overline{FA}\cong\overline{AC}$), so $\overline{EB}$ intersects $\overline{FC}$ at its midpoint $A$, so $\overline{EB}$ is a segment bisector (bisecting $\overline{FC}$). Wait, yes! Because $\overline{EB}$ intersects $\overline{FC}$ at $A$, which is the midpoint of $\overline{FC}$ (since $FA = AC$), so $\overline{EB}$ bisects $\overline{FC}$, hence $\overline{EB}$ is a segment bisector. So this is true.

- **$FA = \frac{1}{2}FC$**: Since $A$ is midpoint of $\overline{FC}$ (because $\overline{FA}\cong\overline{AC}$), so $FA = AC$, and $FC = FA + AC = 2FA$, so $FA = \frac{1}{2}FC$. True.

- **$\overline{DA}\cong\overline{AB}$**: Is there any marking or given that $DA = AB$? The given is $\overline{ED}\cong\overline{DB}$, but no info about $DA$ and $AB$. So we can't assume that. So false.

- **A is the midpoint of $\overline{FC}$**: As $\overline{FA}\cong\overline{AC}$ (from the tick marks on $\overline{FC}$: $FA$ and $AC$ have same tick), so $A$ is midpoint. True.

Wait, let's re-express with correct analysis:

1. **$\overline{EB}$ is bisected by $\overline{DF}$**: Given $\overline{ED}\cong\overline{DB}$, so $D$ is midpoint of $\overline{EB}$. $\overline{DF}$ passes through $D$, so $\overline{DF}$ bisects $\overline{EB}$ (since it intersects at midpoint $D$). So this is true? Wait, the statement is "$\overline{EB}$ is bisected by $\overline{DF}$" – yes, because $D$ is midpoint, and $\overline{DF}$ goes through $D$.

2. **A is the midpoint of $\overline{FC}$**: The tick marks on $\overline{FC}$ (one on $FA$, one on $AC$) indicate $FA = AC$, so $A$ is midpoint. True.

3. **$\overline{FC}$ bisects $\overline{EB}$**: $\overline{FC}$ intersects $\overline{EB}$ at $A$. For $\overline{FC}$ to bisect $\overline{EB}$, $A$ must be midpoint of $\overline{EB}$, but midpoint is $D$ (since $ED = DB$), so $A
eq D$, so false.

4. **$\overline{EB}$ is a segment bisector**: $\overline{EB}$ intersects $\overline{FC}$ at $A$, which is midpoint of $\overline{FC}$ (since $FA = AC$), so $\overline{EB}$ bisects $\overline{FC}$, hence $\overline{EB}$ is a segment bisector. True.

5. **$FA = \frac{1}{2}FC$**: Since $A$ is midpoint, $FC = FA + AC = 2FA$, so $FA = \frac{1}{2}FC$. True.

6. **$\overline{DA}\cong\overline{AB}$**: No given or marking for $DA = AB$, so false.

Wait, but the first option: "$\overline{EB}$ is bisected by $\overline{DF}$" – let's confirm. The segment $\overline{EB}$ has endpoints $E$ and $B$, midpoint $D$ (since $ED = DB$). The segment $\overline{DF}$ connects $D$ to $F$, so it passes through the midpoint of $\overline{EB}$, hence $\overline{DF}$ bisects $\overline{EB}$, so the statement "$\overline{EB}$ is bisected by $\overline{DF}$" is true.

Now, let's list the true ones:

- $\overline{EB}$ is bisected by $\overline{DF}$: True (because $D$ is midpoint of $\overline{EB}$, and $\overline{DF}$ passes through $D$).

- A is the midpoint of $\overline{FC}$: True (since $FA = AC$ from tick marks).

- $\overline{EB}$ is a segment bisector: True (because $\overline{EB}$ intersects $\overline{FC}$ at $A$, midpoint of $\overline{FC}$).

- $FA = \frac{1}{2}FC$: True (since $A$ is midpoint, $FC = 2FA$).

Wait, but maybe I made a mistake with the first option. Let's re-examine the figure: $D$ is on $\overline{EB}$, $A$ is on $\overline{EB}$ to the right of $D$. The given is $\overline{ED}\cong\overline{DB}$, so $ED = DB$, so $D$ is midpoint of $\overline{EB}$. Then $\overline{DF}$ is a segment from $D$ to $F$, so $\overline{DF}$ intersects $\overline{EB}$ at $D$, its midpoint, so $\overline{DF}$ bisects $\overline{EB}$, which means $\overline{EB}$ is bisected by $\overline{DF}$. So that's true.

Now, the option "A is the midpoint of $\overline{FC}$": the tick marks on $\overline{FC}$ (one on $FA$, one on $AC$) mean $FA = AC$, so $A$ is midpoint. True.

"$\overline{EB}$ is a segment bisector": $\overline{EB}$ intersects $\overline{FC}$ at $A$, which is midpoint of $\overline{FC}$, so $\overline{EB}$ bisects $\overline{FC}$, hence $\overline{EB}$ is a segment bisector. True.

"$FA = \frac{1}{2}FC$": Since $A$ is midpoint, $FC = FA + AC = 2FA$, so $FA = \frac{1}{2}FC$. True.

Wait, but what about "A is the midpoint of $\overline{FC}$" – yes, that's true.

Wait, maybe the initial given is $\overline{ED}\cong\overline{DB}$ (so $D$ is midpoint of $\overline{EB}$) and the tick marks on $\overline{FC}$ (so $A$ is midpoint of $\overline{FC}$). Let's recheck each option:

1. $\overline{EB}$ is bisected by $\overline{DF}$: $\overline{EB}$'s midpoint is $D$, and $\overline{DF}$ passes through $D$, so yes, $\overline{DF}$ bisects $\overline{EB}$, so the statement is true.

2. A is the midpoint of $\overline{FC}$: Tick marks on $\overline{FA}$ and $\overline{AC}$ mean $FA = AC$, so $A$ is midpoint. True.

3. $\overline{FC}$ bisects $\overline{EB}$: $\overline{FC}$ intersects $\overline{EB}$ at $A$, but $\overline{EB}$'s midpoint is $D
eq A$, so no. False.

4. $\overline{EB}$ is a segment bisector: $\overline{EB}$ intersects $\overline{FC}$ at $A$, midpoint of $\overline{FC}$, so $\overline{EB}$ bisects $\overline{FC}$, hence it's a segment bisector. True.

5. $FA = \frac{1}{2}FC$: Since $A$ is midpoint, $FC = 2FA \implies FA = \frac{1}{2}FC$. True.

6. $\overline{DA}\cong\overline{AB}$: No info, so false.

So the true statements are:

- $\overline{EB}$ is bisected by $\overline{DF}$

- A is the midpoint of $\overline{FC}$

- $\overline{EB}$ is a segment bisector

- $FA = \frac{1}{2}FC$

Wait, but let's confirm with the problem's given: "Given $\overline{ED}\cong\overline{DB}$" (so $ED = DB$) and the tick marks on $\overline{FA}$ and $\overline{AC}$ (so $FA = AC$).

Now, let's check each option again:

1. $\overline{EB}$ is bisected by $\overline{DF}$: Correct, because $D$ is midpoint of $\overline{EB}$, and $\overline{DF}$ passes through $D$.

2. A is the midpoint of $\overline{FC}$: Correct, $FA = AC$.

3. $\overline{FC}$ bisects $\overline{EB}$: Incorrect, since $\overline{FC}$ intersects $\overline{EB}$ at $A$, not midpoint $D$.

4. $\overline{EB}$ is a segment bisector: Correct, because $\overline{EB}$ intersects $\overline{FC}$ at its midpoint $A$, so it bisects $\overline{FC}$.

5. $FA = \frac{1}{2}FC$: Correct, $FC = FA + AC = 2FA \implies FA = \frac{FC}{2}$.

6. $\overline{DA}\cong\overline{AB}$: Incorrect, no info.

So the true options are:

- $\overline{EB}$ is bisected by $\overline{DF}$

- A is the midpoint of $\overline{FC}$

- $\overline{EB}$ is a segment bisector

- $FA = \frac{1}{2}FC$

Wait, but maybe the first option is not true. Wait, the segment being bisected is $\overline{EB}$, and the bisector is $\overline{DF}$. For a segment to be bisected by another segment, the bisector must intersect it at its midpoint. $\overline{DF}$ intersects $\overline{EB}$ at $D$, which is midpoint (since $ED = DB$), so yes, $\overline{DF}$ bisects $\overline{EB}$, so the statement "$\overline{EB}$ is bisected by $\overline{DF}$" is true.

# Answer:
- $\overline{EB}$ is bisected by $\overline{DF}$
- A is the midpoint of $\overline{FC}$
- $\overline{EB}$ is a segment bisector
- $FA = \frac{1}{2}FC$

(Note: The options with these statements are the correct ones. In the original problem's checkbox format, the true options are:

- $\overline{EB}$ is bisected by $\overline{DF}$ (checkbox checked)
- A is the midpoint of $\overline{FC}$ (checkbox checked)
- $\overline{EB}$ is a segment bisector (checkbox checked)
- $FA = \frac{1}{2}FC$ (checkbox checked)

The other options are false.)

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