Question

h is the circumcenter, or point of concurrency, of the perpendicular bisectors of δace. which statements must be true regarding the diagram? □ ∠hdc ≅ ∠hde □ \\(\overline{ab} \cong \overline{af}\\) □ \\(\overline{ab} \cong \overline{bc}\\) □ \\(\overline{hc} \cong \overline{he}\\) □ \\(\overline{ha} \cong \overline{hb}\\)

Explanation:

Step1: Analyze ∠HDC ≅ ∠HDE

Since HD is a perpendicular bisector (as H is circumcenter, so perpendicular bisectors meet at H), ∠HDC and ∠HDE are both right angles (90°), so they are congruent. So this statement is true.

Step2: Analyze \(\overline{AB} \cong \overline{AF}\)

AB is on the perpendicular bisector of AC (since B is right angle), so AB = BC? Wait, no, AB is a segment from A to B (on AC's perpendicular bisector), and AF is on AE's perpendicular bisector. Wait, actually, B is the midpoint? Wait, no, perpendicular bisector: if HB is perpendicular to AC, then AB = BC? Wait, no, perpendicular bisector of a segment: the point on the perpendicular bisector is equidistant from the endpoints. Wait, H is circumcenter, so HB is perpendicular bisector of AC, so AB = BC? Wait, no, HB is perpendicular to AC, so B is the midpoint of AC? Wait, yes! Because perpendicular bisector of AC: so B is midpoint, so AB = BC? Wait, no, AB is from A to B (midpoint), so AB = BC? Wait, no, AC is the segment, so AB = BC (since B is midpoint). Wait, but the option is \(\overline{AB} \cong \overline{AF}\). Wait, AF: F is on AE, and HF is perpendicular bisector of AE, so F is midpoint of AE, so AF = FE. But AB is midpoint of AC, so AB = BC, but is AB = AF? Not necessarily, unless AC = AE, which isn't given. Wait, maybe I made a mistake. Wait, HB is perpendicular bisector of AC, so AB = BC (since B is midpoint). HF is perpendicular bisector of AE, so AF = FE. But AB and AF: unless AC = AE, they aren't equal. So this statement is not necessarily true.

Step3: Analyze \(\overline{AB} \cong \overline{BC}\)

Since HB is the perpendicular bisector of AC (because H is circumcenter, so HB is perpendicular to AC and B is midpoint of AC), so AB = BC (midpoint divides AC into two equal parts). Wait, yes! Because B is midpoint of AC, so AB = BC. Wait, but earlier I thought AB = BC, but let's confirm: perpendicular bisector of AC: so HB ⊥ AC and AB = BC (since B is midpoint). So this statement is true? Wait, no, wait: AB is from A to B (midpoint), so AB = BC (length of AC is AB + BC, so AB = BC). So \(\overline{AB} \cong \overline{BC}\) is true? Wait, but let's check other options.

Step4: Analyze \(\overline{HC} \cong \overline{HE}\)

H is the circumcenter, so it is equidistant from all vertices of the triangle. So HC = HE = HA (circumradius). So \(\overline{HC} \cong \overline{HE}\) is true, because H is equidistant from C and E (since H is circumcenter, lies on perpendicular bisectors of all sides, so distance from H to C and H to E is equal (circumradius)).

Step5: Analyze \(\overline{HA} \cong \overline{HB}\)

HA is the circumradius (distance from H to A), HB is the distance from H to B. But B is on AC, so HB is the length of the perpendicular from H to AC, while HA is the circumradius. Unless the triangle is isoceles, HA ≠ HB. So this is not true.

Wait, let's re-examine:

- ∠HDC ≅ ∠HDE: both 90°, so true.
- \(\overline{AB} \cong \overline{AF}\): AB is midpoint of AC, AF is midpoint of AE. AC and AE are sides of triangle, not necessarily equal, so not true.
- \(\overline{AB} \cong \overline{BC}\): B is midpoint of AC (since HB is perpendicular bisector), so AB = BC. So this is true? Wait, but earlier I thought AB = BC, but let's check the diagram. HB is perpendicular to AC, so B is midpoint, so AB = BC. So this statement is true? Wait, but the option is \(\overline{AB} \cong \overline{BC}\), so yes, because B is midpoint.
- \(\overline{HC} \cong \overline{HE}\): H is circumcenter, so HC = HE (circumradius), true.
- \(\overline{HA} \cong \overline{HB}\): HA is circumradius, HB is distance from H to B (on AC), not equal, so false.

Wait, but maybe I messed up AB and BC. Wait, AC is the segment, B is on AC, and HB is perpendicular to AC, so B is midpoint, so AB = BC. So \(\overline{AB} \cong \overline{BC}\) is true. But also ∠HDC ≅ ∠HDE (both 90°), and \(\overline{HC} \cong \overline{HE}\) (circumradius). Wait, let's check again.

Wait, circumcenter: the point where perpendicular bisectors meet, so H is equidistant from A, C, E. So HA = HC = HE. So \(\overline{HC} \cong \overline{HE}\) is true (since HC = HE). ∠HDC and ∠HDE: HD is perpendicular to CE (since H is circumcenter, so HD is perpendicular bisector of CE), so ∠HDC and ∠HDE are both 90°, so congruent. \(\overline{AB} \cong \overline{BC}\): since HB is perpendicular bisector of AC, B is midpoint, so AB = BC. Wait, but is AB = BC? Yes, because B is midpoint of AC. So that's true. Wait, but the options: let's list the true ones.

Wait, maybe I made a mistake with \(\overline{AB} \cong \overline{AF}\). AF: F is midpoint of AE (since HF is perpendicular bisector), so AF = FE. AB is midpoint of AC, so AB = BC. So AB and AF: unless AC = AE, not equal. So \(\overline{AB} \cong \overline{AF}\) is false. \(\overline{HA} \cong \overline{HB}\): HA is circumradius, HB is distance from H to B (on AC), not equal, so false.

So the true statements are:

- ∠HDC ≅ ∠HDE (both 90°)
- \(\overline{AB} \cong \overline{BC}\) (B is midpoint of AC)
- \(\overline{HC} \cong \overline{HE}\) (H is equidistant from C and E)

Wait, but let's check the diagram again. HB is perpendicular to AC, so B is midpoint, so AB = BC. HD is perpendicular to CE, so D is midpoint, so CD = DE. HF is perpendicular to AE, so F is midpoint, so AF = FE. H is circumcenter, so HA = HC = HE. So:

- ∠HDC ≅ ∠HDE: both right angles, true.
- \(\overline{AB} \cong \overline{AF}\): AB is midpoint of AC, AF midpoint of AE. AC and AE not necessarily equal, so false.
- \(\overline{AB} \cong \overline{BC}\): AB = BC (midpoint), true.
- \(\overline{HC} \cong \overline{HE}\): HC = HE (circumradius), true.
- \(\overline{HA} \cong \overline{HB}\): HA > HB (since HA is circumradius, HB is segment from H to B on AC), false.

Wait, but maybe the problem is that AB is not necessarily equal to BC? Wait, no, perpendicular bisector of AC: the line HB is the perpendicular bisector, so any point on HB is equidistant from A and C. But B is on AC, so B must be the midpoint, because HB is perpendicular to AC. So AB = BC. Yes, that's a theorem: the perpendicular bisector of a segment intersects the segment at its midpoint. So B is midpoint of AC, so AB = BC.

So the true statements are ∠HDC ≅ ∠HDE, \(\overline{AB} \cong \overline{BC}\), and \(\overline{HC} \cong \overline{HE}\). Wait, but let's check the options again.

Wait, the options are:

1. ∠HDC ≅ ∠HDE: true (both 90°)
2. \(\overline{AB} \cong \overline{AF}\): false
3. \(\overline{AB} \cong \overline{BC}\): true (B is midpoint)
4. \(\overline{HC} \cong \overline{HE}\): true (H is circumcenter, so HC = HE)
5. \(\overline{HA} \cong \overline{HB}\): false

But maybe I made a mistake with \(\overline{AB} \cong \overline{BC}\). Wait, AC is the side, B is on AC, and HB is perpendicular to AC, so B is the midpoint, so AB = BC. Yes. So that's true.

Wait, but let's confirm with circumcenter properties. Circumcenter is equidistant from all vertices, so HA = HC = HE. So \(\overline{HC} \cong \overline{HE}\) is true. ∠HDC and ∠HDE: HD is perpendicular to CE, so both angles are 90°, so congruent. \(\overline{AB} \cong \overline{BC}\): since HB is perpendicular bisector of AC, B is midpoint, so AB = BC. So these three are true. But maybe the problem considers some of them. Wait, maybe the intended answers are ∠HDC ≅ ∠HDE, \(\overline{HC} \cong \overline{HE}\), and maybe \(\overline{AB} \cong \overline{BC}\). But let's check again.

Alternatively, maybe \(\overline{AB} \cong \overline{BC}\) is not true. Wait, no: if HB is the perpendicular bisector of AC, then B is the midpoint of AC, so AB = BC. Yes, that's a basic property of perpendicular bisectors: they intersect the segment at its midpoint. So AB = BC. So that's true.

So the true statements are:

- ∠HDC ≅ ∠HDE
- \(\overline{AB} \cong \overline{BC}\)
- \(\overline{HC} \cong \overline{HE}\)

But let's check the options again. The options are:

1. ∠HDC ≅ ∠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 these three are true. But maybe the problem has multiple correct answers. Let's confirm each:

1. ∠HDC ≅ ∠HDE: HD is perpendicular to CE (since H is circumcenter, so HD is perpendicular bisector), so both angles are 90°, so congruent. True.

2. \(\overline{AB} \cong \overline{AF}\): AB is midpoint of AC, AF midpoint of AE. No info AC = AE, so false.

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

4. \(\overline{HC} \cong \overline{HE}\): H is circumcenter, so HC = HE (equidistant from C and E). True.

5. \(\overline{HA} \cong \overline{HB}\): HA is circumradius, HB is distance from H to B (on AC), not equal. False.

So the correct options are the first, third, and fourth. But maybe the problem expects these.

Answer:

The true statements are:

• \( \angle \text{HDC} \cong \angle \text{HDE} \)
• \( \overline{\text{AB}} \cong \overline{\text{BC}} \)
• \( \overline{\text{HC}} \cong \overline{\text{HE}} \)

(To mark the checkboxes: check \( \angle \text{HDC} \cong \angle \text{HDE} \), \( \overline{\text{AB}} \cong \overline{\text{BC}} \), and \( \overline{\text{HC}} \cong \overline{\text{HE}} \).)