h is the circumcenter, or point of concurrency, of the perpendicular bisectors of $ \triangle ace $. which statements must be true regarding the diagram? $ \angle hdc \cong \angle hde $ $ \overline{ab} \cong \overline{af} $ $ \overline{ab} \cong \overline{bc} $ $ \overline{hc} \cong \overline{he} $ $ \overline{ha} \cong \overline{hb} $

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# Brief Explanations:
1. For $\angle HDC \cong \angle HDE$: Since $HD$ is a perpendicular bisector (as $H$ is circumcenter, perpendicular bisectors meet at $H$), $\angle HDC$ and $\angle HDE$ are right angles (both $90^\circ$), so they are congruent.
2. For $\overline{AB} \cong \overline{AF}$: There's no information suggesting $AB$ and $AF$ are related by bisecting or equal length. $B$ is on $AC$, $F$ is on $AE$, no reason for them to be equal.
3. For $\overline{AB} \cong \overline{BC}$: $B$ is the foot of perpendicular from $H$ to $AC$, so $AB = BC$ (perpendicular bisector of $AC$: $HB$ bisects $AC$), so this is true? Wait, no—wait, $HB$ is perpendicular bisector of $AC$, so $AB = BC$. Wait, but let's recheck. Wait, the diagram: $B$ is on $AC$, $HB \perp AC$, so $HB$ is perpendicular bisector, so $AB = BC$. Wait, but also, $HC = HE$: circumcenter is equidistant from all vertices, so $HC = HE = HA$, since $H$ is circumcenter (distance from circumcenter to vertices is circumradius). So $HC \cong HE$ is true. And $\angle HDC \cong \angle HDE$: both right angles, true. $\overline{AB} \cong \overline{BC}$: since $HB$ is perpendicular bisector of $AC$, $AB = BC$, true? Wait, maybe I made a mistake. Wait, let's analyze each:

- $\angle HDC \cong \angle HDE$: $HD \perp CE$ (since $D$ is foot of perpendicular bisector), so $\angle HDC = \angle HDE = 90^\circ$, so congruent. True.
- $\overline{AB} \cong \overline{AF}$: No, $AB$ is part of $AC$, $AF$ part of $AE$, no relation. False.
- $\overline{AB} \cong \overline{BC}$: $HB$ is perpendicular bisector of $AC$, so $AB = BC$. True? Wait, perpendicular bisector of a segment divides it into two equal parts, so yes, $AB = BC$.
- $\overline{HC} \cong \overline{HE}$: $H$ is circumcenter, so distance from $H$ to $C$ and $H$ to $E$ is circumradius, so $HC = HE$. True.
- $\overline{HA} \cong \overline{HB}$: $HA$ is circumradius (distance from $H$ to $A$), $HB$ is distance from $H$ to $B$ (a point on $AC$), not necessarily equal. False.

Wait, but maybe my initial thought about $\overline{AB} \cong \overline{BC}$ is correct because $HB$ is perpendicular bisector. But let's confirm the circumcenter properties: circumcenter is intersection of perpendicular bisectors, so each perpendicular bisector bisects a side and is perpendicular to it. So $HB$ bisects $AC$ (so $AB = BC$), $HD$ bisects $CE$ (so $CD = DE$), $HF$ bisects $AE$ (so $AF = FE$). Also, $HA = HC = HE$ (circumradius). So:

- $\angle HDC \cong \angle HDE$: both $90^\circ$, true.
- $\overline{AB} \cong \overline{BC}$: $HB$ bisects $AC$, so $AB = BC$, true.
- $\overline{HC} \cong \overline{HE}$: $HA = HC = HE$, true.

Wait, but the options: let's check again. Wait, maybe I messed up $\overline{AB} \cong \overline{BC}$. Wait, $B$ is on $AC$, $HB \perp AC$, so $HB$ is perpendicular bisector, so $AB = BC$. Yes. And $\angle HDC \cong \angle HDE$: both right angles, true. $\overline{HC} \cong \overline{HE}$: true. Wait, but the original options: let's list again:

Options:

1. $\angle HDC \cong \angle HDE$: True (both $90^\circ$)
2. $\overline{AB} \cong \overline{AF}$: False
3. $\overline{AB} \cong \overline{BC}$: True (perpendicular bisector of $AC$)
4. $\overline{HC} \cong \overline{HE}$: True (circumradius)
5. $\overline{HA} \cong \overline{HB}$: False (HA is circumradius, HB is distance to midpoint, not equal)

Wait, but maybe the diagram has $B$ on $AC$, $D$ on $CE$, $F$ on $AE$. So $HB \perp AC$, so $AB = BC$ (since $HB$ is perpendicular bisector). $HD \perp CE$, so $CD = DE$, and $\angle HDC = \angle HDE = 90^\circ$. $H$ is circumcenter, so $HA = HC = HE$. So the true statements are $\angle HDC \cong \angle HDE$, $\overline{AB} \cong \overline{BC}$, $\overline{HC} \cong \overline{HE}$? Wait, but maybe I made a mistake with $\overline{AB} \cong \overline{BC}$. Wait, perpendicular bisector of $AC$: the perpendicular bisector of a segment is the set of points equidistant from $A$ and $C$. $H$ is on it, and $B$ is the foot, so $AB = BC$. Yes. So those three? But let's check the options again. Wait, the user's question is "Which statements must be true", so we need to identify all true ones.

Wait, let's re-express:

- $\angle HDC \cong \angle HDE$: True (both right angles, $HD \perp CE$)
- $\overline{AB} \cong \overline{AF}$: False (no relation)
- $\overline{AB} \cong \overline{BC}$: True (HB is perpendicular bisector of AC, so AB = BC)
- $\overline{HC} \cong \overline{HE}$: True (H is circumcenter, so HC = HE = HA)
- $\overline{HA} \cong \overline{HB}$: False (HA is radius, HB is distance to AC, not equal)

But maybe the intended answers are $\angle HDC \cong \angle HDE$, $\overline{HC} \cong \overline{HE}$, and $\overline{AB} \cong \overline{BC}$? Wait, but let's check the diagram again. Wait, maybe $B$ is not the midpoint? Wait, no—circumcenter is the intersection of perpendicular bisectors, so each perpendicular bisector passes through the midpoint of a side and is perpendicular to it. So $HB$ is perpendicular bisector of $AC$, so $B$ is midpoint of $AC$, so $AB = BC$. Yes. So that's true.

So the true statements are:

- $\angle HDC \cong \angle HDE$
- $\overline{AB} \cong \overline{BC}$
- $\overline{HC} \cong \overline{HE}$

Wait, but maybe I was wrong about $\overline{AB} \cong \overline{BC}$. Wait, let's take a step back. The circumcenter is the intersection of perpendicular bisectors, so each perpendicular bisector:

- For side $AC$: the perpendicular bisector is the line through $H$ and $B$ (since $HB \perp AC$), so $B$ is midpoint of $AC$, so $AB = BC$. Correct.

- For side $CE$: perpendicular bisector is line through $H$ and $D$, so $D$ is midpoint of $CE$, $CD = DE$, and $\angle HDC = \angle HDE = 90^\circ$, so congruent.

- For side $AE$: perpendicular bisector is line through $H$ and $F$, so $F$ is midpoint of $AE$, $AF = FE$.

And $HA = HC = HE$ (circumradius), so $HC = HE$ is true.

So the true statements are $\angle HDC \cong \angle HDE$, $\overline{AB} \cong \overline{BC}$, $\overline{HC} \cong \overline{HE}$. But let's check the options again. Wait, maybe the original problem's diagram has $B$ not as midpoint? No, because $HB$ is perpendicular bisector, so it must bisect $AC$. So $AB = BC$.

But maybe the intended answers are $\angle HDC \cong \angle HDE$, $\overline{HC} \cong \overline{HE}$, and $\overline{AB} \cong \overline{BC}$? Wait, but let's confirm with definitions.

Circumcenter: the point where perpendicular bisectors of a triangle meet, and it is equidistant from all three vertices (so $HA = HC = HE$).

Perpendicular bisector of a segment: a line that is perpendicular to the segment and bisects it (so midpoint, and right angle).

So:

- $\angle HDC \cong \angle HDE$: Both are right angles (since $HD$ is perpendicular to $CE$), so congruent. True.

- $\overline{AB} \cong \overline{AF}$: No, $AB$ is part of $AC$, $AF$ part of $AE$, no midpoint or bisecting relation. False.

- $\overline{AB} \cong \overline{BC}$: $B$ is midpoint of $AC$ (since $HB$ is perpendicular bisector), so $AB = BC$. True.

- $\overline{HC} \cong \overline{HE}$: $HC$ and $HE$ are both circumradii (distance from $H$ to vertices $C$ and $E$), so equal. True.

- $\overline{HA} \cong \overline{HB}$: $HA$ is circumradius, $HB$ is distance from $H$ to $AC$ (a side), not equal (unless triangle is isoceles in a way, but not necessarily). False.

So the true statements are $\angle HDC \cong \angle HDE$, $\overline{AB} \cong \overline{BC}$, $\overline{HC} \cong \overline{HE}$. But let's check the options again. Wait, maybe the user's diagram has $B$ not as midpoint? No, because $HB$ is perpendicular bisector, so it must bisect $AC$. So $AB = BC$.

But maybe the intended answers are $\angle HDC \cong \angle HDE$, $\overline{HC} \cong \overline{HE}$, and $\overline{AB} \cong \overline{BC}$? Wait, but let's see the options:

1. $\angle HDC \cong \angle HDE$: True
2. $\overline{AB} \cong \overline{AF}$: False
3. $\overline{AB} \cong \overline{BC}$: True
4. $\overline{HC} \cong \overline{HE}$: True
5. $\overline{HA} \cong \overline{HB}$: False

So those three are true. But maybe the problem expects $\angle HDC \cong \angle HDE$, $\overline{HC} \cong \overline{HE}$, and $\overline{AB} \cong \overline{BC}$? Wait, but let's confirm with the first option: $\angle HDC$ and $\angle HDE$ are both right angles, so congruent. Correct. $\overline{HC} \cong \overline{HE}$: circumradius, correct. $\overline{AB} \cong \overline{BC}$: midpoint, correct.

# Answer:
- $\boldsymbol{\angle HDC \cong \angle HDE}$ (True, both right angles)
- $\boldsymbol{\overline{AB} \cong \overline{BC}}$ (True, $HB$ bisects $AC$)
- $\boldsymbol{\overline{HC} \cong \overline{HE}}$ (True, circumradius)

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