Explanation:
Okay, so I have this geometry problem here. Let me try to figure it out step by step. The problem says:

Altitudes AD and BE of triangle ABC intersect at H. If angle BAC is 54 degrees and angle ABC is 52 degrees, then what is angle AHB?

Alright, let me start by drawing triangle ABC. Let me recall that in any triangle, the altitudes intersect at a point called the orthocenter. So here, H is the orthocenter because AD and BE are altitudes intersecting at H.

First, maybe I should sketch triangle ABC with the given angles. Let me note down the given angles:

Angle BAC (which is angle A) is 54 degrees, and angle ABC (which is angle B) is 52 degrees. So angle C can be found since the sum of angles in a triangle is 180 degrees. Let me calculate angle C first.

So angle C = 180° - angle A - angle B = 180° - 54° - 52°. Let me compute that. 54 + 52 is 106, so 180 - 106 is 74. So angle C is 74 degrees. Got that.

Now, the problem is about angle AHB. Let me recall that H is the orthocenter. So angle AHB is the angle between the altitudes AD and BE. Hmm. How can I find this angle?

I need to recall some properties of orthocenters and maybe some relationships between the angles. Let me think. In triangle ABC, the orthocenter H has some relations with the original angles. Maybe I can use the fact that in certain configurations, angles formed at the orthocenter relate to the original angles.

Alternatively, maybe I can find some cyclic quadrilaterals here? Because altitudes create right angles with the sides. Let me think.

Since AD is an altitude from A to BC, that means AD is perpendicular to BC. Similarly, BE is an altitude from B to AC, so BE is perpendicular to AC. Therefore, AD ⊥ BC and BE ⊥ AC.

Now, let me consider quadrilateral formed by H, the feet of the altitudes, and other points. Wait, but maybe I need to look at triangle AHB. Let me focus on triangle AHB. What angles do I know in this triangle?

I need to find angle at H, which is angle AHB. To find this angle, perhaps I can find other angles in triangle AHB. For example, if I can find angles at A and B in triangle AHB, then I can subtract from 180 degrees to get angle AHB.

But wait, in triangle AHB, the angles at A and B are not the same as the original angles of triangle ABC. Let me think. Let me denote some variables. Let me denote angle HAB and angle HBA. Then angle AHB = 180° - angle HAB - angle HBA.

But what are angle HAB and angle HBA?

Since AD is the altitude from A to BC, then AD is perpendicular to BC, so angle ADC is 90°, and angle ADB is 90°. Similarly, BE is perpendicular to AC, so angle BEA is 90°, and angle BEC is 90°.

Let me consider angle HAB. Since AD is an altitude, angle between AD and AB is angle BAD. Wait, angle BAD is part of angle BAC, which is 54 degrees. Let me try to find angle BAD.

Wait, angle BAD is the angle between AB and AD. But since AD is perpendicular to BC, maybe I can find angle BAD using some relations. Let me recall that in triangle ABC, the altitude AD divides the triangle into two right triangles: ABD and ADC.

In triangle ABD, angle ABD is angle ABC, which is 52 degrees. Wait, no. Wait, angle ABD is part of angle ABC. Wait, actually, angle ABD is angle between AB and BD, but BD is part of BC. Hmm. Maybe it's easier to use trigonometry here.

Alternatively, maybe I can use the fact that in triangle ABC, the measure of angle between two altitudes can be related to the original angles. Wait, I remember that in a triangle, the measure of angle AHB is equal to 180° minus angle ACB. Wait, is that correct?

Wait, let me check. Let me recall that in triangle ABC, the orthocenter H creates some angles. For example, in acute triangles, the orthocenter lies inside the triangle. Let me suppose that triangle ABC is acute. Given that angles A and B are 54° and 52°, which are both less than 90°, and angle C is 74°, also less than 90°. So yes, triangle ABC is acute, so H lies inside the triangle.

Now, there's a property that angle AHB = 180° - angle ACB. Wait, if that's the case, then angle AHB would be 180° - 74° = 106°. But let me verify if this is correct. Let me think why that might be.

Alternatively, maybe angle AHB = 180° - angle ABC? Wait, no. Let me think more carefully.

Let me consider quadrilateral formed by H, the feet of the altitudes, and maybe some other points. Wait, let me consider triangles involving H.

Since AD and BE are altitudes, then H lies at their intersection. Let me consider triangle AHB. Let me try to find angles HAB and HBA.

Let me note that angle HAB is equal to angle between AB and altitude AD. Since AD is an altitude, AD is perpendicular to BC. Therefore, angle BAD is equal to 90° minus angle ABC? Wait, no. Wait, angle BAD is part of angle BAC.

Alternatively, maybe I can look at triangle ABD, which is right-angled at D. In triangle ABD, angle at D is 90°, angle at B is angle ABD, which is angle ABC? Wait, no. Wait, angle ABD is part of angle ABC. Wait, angle ABC is angle at B, which is 52°, and since D is on BC, angle ABD is the same as angle ABC? Wait, no. Wait, point D is on BC, so BD is part of BC. Therefore, angle ABD is equal to angle ABC? Wait, no. Wait, angle at B in triangle ABC is angle between AB and BC. But angle ABD is angle between AB and BD. Since BD is part of BC, then angle ABD is the same as angle ABC. Wait, that seems incorrect because BD is just a segment along BC, so angle between AB and BD is the same as angle between AB and BC, which is angle ABC. Wait, is that true?

Wait, suppose I have triangle ABC, with point D on BC. Then angle ABD is the same as angle ABC if D is on BC. Wait, no. Wait, if D is on BC, then BD is just a segment from B to D on BC. So angle between AB and BD is the same as angle between AB and BC, which is angle ABC. Wait, but that seems to suggest angle ABD is equal to angle ABC, but that's only if D is at point C. Wait, no, if D is somewhere along BC, then angle ABD is still part of angle ABC. Wait, maybe I need to think more carefully.

Let me take coordinates. Maybe assigning coordinates to triangle ABC could help. Alternatively, maybe using trigonometric relations.

Alternatively, maybe using the fact that in triangle ABC, angles at H can be found by considering that the orthocenter creates similar triangles or something.

Alternatively, here's a method I remember: in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Wait, let me check with an example. Suppose triangle ABC is equilateral, all angles 60°, then orthocenter is also the centroid and incenter. Then angle AHB would be... Well, in that case, angle AHB would be 120°, since each angle at the orthocenter in an equilateral triangle is 120°, which is 180° - 60°, so yes, 180° - angle C. So if angle C is 60°, angle AHB is 120°, which matches. Similarly, maybe in general angle AHB = 180° - angle ACB?

Wait, if that's the case, then in our problem, angle ACB is 74°, so angle AHB would be 180° - 74° = 106°, which would be the answer. But I need to verify this.

Alternatively, let me try to derive it. Let me think about quadrilateral formed by the orthocenter and some points.

Since AD and BE are altitudes, intersecting at H. So, let me consider quadrilateral AEHD, where E is the foot of BE on AC, and D is the foot of AD on BC. Wait, maybe not. Alternatively, let me consider angles at H.

Since BE is an altitude, BE ⊥ AC. So angle BEA is 90°, and similarly AD ⊥ BC, so angle ADB is 90°. Let me consider triangle AHB. Let me think about angles in triangle AHB.

Alternatively, since BE is perpendicular to AC, then angle between BE and AC is 90°, so angle between BE and AB can be calculated. Wait, angle between BE and AB is angle ABE. Similarly, angle between AD and AB is angle BAD. Then, maybe angle HAB is equal to angle BAD?

Wait, angle BAD is part of angle BAC. Let me think. Let me try to find angle HAB and angle HBA.

Let me denote angle HAB as x and angle HBA as y. Then angle AHB = 180° - x - y.

So if I can find x and y, I can find angle AHB.

Now, what is angle HAB? HAB is the angle between HA and AB. But HA is part of the altitude AD. Therefore, angle HAB is equal to angle between altitude AD and AB. Since AD is perpendicular to BC, maybe angle HAB relates to angle ABC?

Wait, let me recall that in triangle ABD, which is right-angled at D, angle at D is 90°, angle at B is angle ABD, and angle at A is angle BAD.

So in triangle ABD: angle ABD + angle BAD + 90° = 180°, so angle ABD + angle BAD = 90°.

But angle ABD is part of angle ABC. Since ABC is 52°, angle ABD is equal to angle ABC? No, wait. If D is on BC, then angle ABD is the same as angle ABC? Wait, no, unless D is at C. Wait, D is the foot of altitude from A to BC. So unless the triangle is right-angled at B, which it is not. Wait, in triangle ABC, angle at B is 52°, and altitude from A to BC is AD. Therefore, BD is a segment of BC, and angle ABD is part of angle ABC.

Wait, let me denote angle ABD as something. Let me think. Let me consider triangle ABC with altitude AD. Then angle BAD = angle BAC - angle DAC. Wait, but maybe I need another approach.

Alternatively, since in triangle ABD, angle BAD + angle ABD = 90°, so angle BAD = 90° - angle ABD.

Similarly, angle ABD is equal to angle ABC, but no, because D is not necessarily C. Wait, angle ABD is equal to angle ABC only if D coincides with C. But D is the foot of altitude from A to BC. Therefore, BD can be found using trigonometry.

Alternatively, maybe I can use trigonometric ratios. Let me assign some lengths. Let me assume BC is of some length, say, let me assign BC = 1 for simplicity. Wait, but maybe that's too vague. Alternatively, maybe assign AB, BC, AC using the given angles?

Alternatively, let me use the fact that in triangle ABC, the altitude from A to BC is AD. Then, the length of AD can be expressed in terms of the sides and angles, but maybe this is overcomplicating.

Wait, maybe I can find angle BAD. Since angle BAD + angle ABD = 90°, and angle ABD is part of angle ABC. Let me denote angle ABD as θ, so angle BAD = 90° - θ. But angle ABC is 52°, so θ is part of it. Wait, but unless I know something else about θ, I can't determine it.

Alternatively, maybe considering triangle ABE. Since BE is an altitude, it is perpendicular to AC. So angle AEB is 90°. In triangle ABE, angle at E is 90°, angle at B is angle ABE, and angle at A is angle BAE.

Similarly, angle BAE is part of angle BAC which is 54°, so angle BAE = angle BAC - angle EAC? Wait, maybe not. Let me think. If BE is altitude from B to AC, then E is on AC, so angle ABE is part of angle ABC?

Wait, this seems getting too tangled. Maybe there is a better approach.

Wait, let me recall that in any triangle, the measure of angle AHB is equal to 180° minus the measure of angle ACB. Wait, if that is true, then, as I thought earlier, angle AHB = 180° - 74° = 106°, so the answer would be 106 degrees.

But why is angle AHB equal to 180° - angle C?

Let me try to derive this. Let me consider quadrilateral HEBD. Wait, but BE and AD are altitudes intersecting at H. Since BE ⊥ AC and AD ⊥ BC, then HE is part of BE, and HD is part of AD. Therefore, angles at E and D are right angles. Therefore, quadrilateral HEBD has two right angles at E and D. But unless other angles are known, not sure.

Alternatively, consider triangle AHB. Let me find the measure of angle AHB by considering the cyclic quadrilaterals.

Wait, here's a better idea. In triangle ABC, the orthocenter H has the property that angles formed at H relate to the original triangle's angles.

For instance, in triangle ABC, the angles at H can be calculated by:

angle AHB = 180° - angle ACB

angle BHC = 180° - angle BAC

angle AHC = 180° - angle ABC

Is this a valid formula?

Let me check with the equilateral triangle case. If all angles are 60°, then angle AHB should be 180° - 60° = 120°, which is correct. If we take a right-angled triangle, say with angle C = 90°, then angle AHB = 180° - 90° = 90°, but in that case, the orthocenter is at point C, so angle AHB would be angle ACB, which is 90°, so that also matches.

Wait, but in the case of a right-angled triangle, orthocenter is at the right-angled vertex, so angle AHB would be angle ACB. Wait, but if angle C is 90°, then angle AHB is 90°, which is equal to angle ACB. However, according to the formula angle AHB = 180° - angle ACB, that would be 180° - 90° = 90°, which matches. So it seems correct.

Therefore, in general, angle AHB = 180° - angle ACB. Therefore, in our problem, angle AHB = 180° - angle C = 180° - 74° = 106°, so the answer is 106 degrees.

But let me check this formula again because sometimes I might mix up the angles.

Alternatively, here's another way to think about it. In triangle ABC, let's consider the orthocenter H. Then, quadrilateral HECD is cyclic? Wait, maybe not. Alternatively, since AD and BE are altitudes, then angles AHB, let me consider the following.

Since BE is perpendicular to AC, then angle between BE and AC is 90°, and AD is perpendicular to BC, angle between AD and BC is 90°. So, in triangle AHB, angles at H.

Wait, if I can find some relationship between angles. Let me think about the fact that in triangle ABC, the angles formed by the altitudes can be related to the original angles.

Alternatively, using the fact that angle between two altitudes is equal to 180° minus the original angle. Wait, if I consider the angles in triangle AHB:

Let me consider angle HAB. Since AD is an altitude, angle BAD = ?

Similarly, angle HBA = ?

Earlier, I thought of using triangle ABD. In triangle ABD, right-angled at D, angle BAD = 90° - angle ABD.

Similarly, angle ABD is angle between AB and BD, but BD is part of BC. Since angle ABC is 52°, then angle between AB and BC is 52°, so angle ABD is some portion of that? Wait, if we can find angle ABD?

Wait, perhaps using the fact that in triangle ABC, the altitude from A divides BC into segments. The ratio of BD to DC can be found using trigonometry? Maybe.

Let me consider triangle ABC, with angle A = 54°, angle B = 52°, angle C = 74°, and let me denote sides as per standard notation: BC = a, AC = b, AB = c.

By the Law of Sines, a / sin A = b / sin B = c / sin C.

So:

a / sin 54° = b / sin 52° = c / sin 74°.

Let me assign some lengths. Let me assume that the length of BC is a = 1 (for simplicity). Then:

a = 1 = BC

Therefore, from Law of Sines:

sin 54° / 1 = sin 52° / b => b = sin 52° / sin 54°

Similarly, c = sin 74° / sin 54°

Let me compute approximate values to see:

First, sin 54° ≈ 0.8090

sin 52° ≈ 0.7880

sin 74° ≈ 0.9613

Therefore:

b ≈ 0.7880 / 0.8090 ≈ 0.9738

c ≈ 0.9613 / 0.8090 ≈ 1.188

So sides:

BC = a = 1

AC = b ≈ 0.9738

AB = c ≈ 1.188

Now, let me find coordinates of the triangle. Let me place point B at (0,0), point C at (1,0). Then I need to find coordinates of point A.

Given that AB ≈ 1.188 and AC ≈ 0.9738. Let me denote coordinates of A as (x, y). Then distance from A to B is sqrt(x² + y²) ≈ 1.188, and distance from A to C is sqrt((x - 1)² + y²) ≈ 0.9738.

So:

x² + y² ≈ (1.188)² ≈ 1.411

(x - 1)² + y² ≈ (0.9738)² ≈ 0.9483

Subtracting the second equation from the first:

x² + y² - [(x - 1)² + y²] ≈ 1.411 - 0.9483

Expanding:

x² - (x² - 2x + 1) ≈ 0.4627

Simplify:

x² - x² + 2x - 1 ≈ 0.4627

So:

2x - 1 ≈ 0.4627 => 2x ≈ 1.4627 => x ≈ 0.7313

Then, plugging back x into x² + y² ≈ 1.411:

(0.7313)^2 + y² ≈ 1.411

Calculating 0.7313² ≈ 0.5348, so y² ≈ 1.411 - 0.5348 ≈ 0.8762 => y ≈ sqrt(0.8762) ≈ 0.936.

So coordinates of A are approximately (0.7313, 0.936). Let me check distance from A to C:

Distance from A(0.7313, 0.936) to C(1,0):

sqrt((1 - 0.7313)^2 + (0 - 0.936)^2) = sqrt((0.2687)^2 + (-0.936)^2) ≈ sqrt(0.0722 + 0.876) ≈ sqrt(0.948) ≈ 0.973, which matches the earlier value of b ≈ 0.9738. Good.

So coordinates:

B(0,0), C(1,0), A(≈0.7313, ≈0.936)

Now, need to find equations of altitudes AD and BE, then find their intersection H, then compute angle AHB.

Let me find altitude AD first. Altitude from A to BC. Since BC is the base from (0,0) to (1,0), which is horizontal. Therefore, the altitude from A to BC is vertical? Wait, no. Wait, BC is horizontal, so its slope is 0, so the altitude from A to BC is vertical if BC is horizontal? Wait, the altitude is perpendicular to BC. Since BC is horizontal, slope 0, then the altitude is vertical, slope undefined. Therefore, the altitude from A is a vertical line passing through A(0.7313, 0.936). Wait, but if BC is horizontal, then the altitude from A to BC is vertical? Wait, no, if BC is horizontal, then the altitude from A would be vertical only if BC is horizontal and the altitude is vertical. Wait, the altitude from A is perpendicular to BC. Since BC is horizontal, the altitude is vertical. Therefore, its equation is x = 0.7313. Therefore, it meets BC at D(0.7313, 0). Therefore, coordinates of D are (0.7313, 0).

Similarly, find equation of altitude BE. BE is the altitude from B to AC. First, find the equation of AC. Points A(0.7313, 0.936) and C(1, 0). Let me compute the slope of AC:

Slope of AC: (0 - 0.936) / (1 - 0.7313) = (-0.936) / (0.2687) ≈ -3.485.

Therefore, the altitude from B(0,0) to AC is perpendicular to AC, so its slope is the negative reciprocal of -3.485, which is 1/3.485 ≈ 0.287.

Therefore, equation of altitude BE is y = 0.287x.

Now, altitude AD is x = 0.7313. Therefore, intersection point H is at x = 0.7313, and y = 0.287 * 0.7313 ≈ 0.287 * 0.7313 ≈ let me compute that:

0.287 * 0.7 = 0.2009, 0.287 * 0.0313 ≈ approx 0.00896. So total ≈ 0.2009 + 0.00896 ≈ 0.2098. So H is approximately at (0.7313, 0.2098).

Now, we need to find angle AHB. Points A(0.7313, 0.936), H(0.7313, 0.2098), B(0,0).

Wait, let me note coordinates:

Point A: (x_A, y_A) ≈ (0.7313, 0.936)

Point H: (x_H, y_H) ≈ (0.7313, 0.2098)

Point B: (0, 0)

So, angle AHB is the angle at point H between points A and B.

Wait, to compute angle AHB, which is the angle at H between HA and HB.

To calculate this angle, I can use coordinates. Let me compute vectors HA and HB, then use the dot product formula.

Vector HA is A - H = (0.7313 - 0.7313, 0.936 - 0.2098) = (0, 0.7262)

Vector HB is B - H = (0 - 0.7313, 0 - 0.2098) = (-0.7313, -0.2098)

Now, angle between vectors HA and HB can be found by:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB: (0)(-0.7313) + (0.7262)(-0.2098) ≈ 0 - 0.7262 * 0.2098 ≈ -0.1524

Compute |HA|: sqrt(0² + 0.7262²) = 0.7262

Compute |HB|: sqrt((-0.7313)^2 + (-0.2098)^2) ≈ sqrt(0.5348 + 0.044) ≈ sqrt(0.5788) ≈ 0.7608

Therefore,

cos(theta) ≈ (-0.1524) / (0.7262 * 0.7608) ≈ (-0.1524) / (0.5525) ≈ -0.2758

Therefore, theta ≈ arccos(-0.2758) ≈ 106 degrees.

Yes! Because arccos(-0.2758) is approximately 106 degrees. Let me check with calculator:

cos(106°): cos(90° + 16°) = -sin(16°) ≈ -0.2756. Which is very close to -0.2758. Therefore, angle theta is approximately 106°, which matches our previous result of 180° - 74° = 106°. Therefore, this confirms that angle AHB is indeed 106 degrees.

Therefore, the answer is 106 degrees.

But let me check once again if there's another way to get this result without coordinates.

Earlier, I thought that angle AHB = 180° - angle ACB. If that formula is correct, then angle AHB = 180° - 74° = 106°, which matches the coordinate calculation.

But why is that formula true? Let me try to derive it.

In triangle ABC, let H be the orthocenter. Then, consider quadrilateral formed by H and the feet of the altitudes. Alternatively, consider the following:

Let me recall that in triangle ABC, angles at H can be related to the original angles. For example, in triangle HBC, angle HBC is equal to 90° - angle ACB. Wait, no, maybe not. Wait, let me think.

Alternatively, consider that in triangle ABC, since BE is an altitude, angle ABE = 90° - angle BAC. Wait, angle ABE is part of angle ABC. Wait, let me try to think step by step.

In triangle ABE, right-angled at E, angle at E is 90°, so angle ABE = 90° - angle BAE. But angle BAE is part of angle BAC, which is 54°, so angle BAE = angle BAC - angle EAC, but I don't know angle EAC.

Wait, this seems complicated. Let me instead consider cyclic quadrilaterals.

Since AD and BE are altitudes, intersecting at H. Then, since AD ⊥ BC and BE ⊥ AC, then points A, B, H, and the feet of the altitudes lie on certain circles?

Alternatively, consider that in triangle ABC, the orthocenter H creates several right angles. For instance, angle AHB is supplementary to angle ACB.

Wait, here's another approach. In triangle ABC, since AD and BE are altitudes, then angles:

In triangle AHB, angle AHB is equal to 180° minus angle HAB minus angle HBA. As I mentioned earlier.

But angle HAB is equal to angle between HA and AB. Since HA is part of altitude AD, which is perpendicular to BC. So angle between AD and AB is angle BAD.

Similarly, angle HBA is equal to angle between HB and BA, which is angle between HB and BA. Since HB is part of altitude BE, which is perpendicular to AC. Therefore, angle between BE and BA is angle EBA.

Therefore, angle HAB = angle BAD, and angle HBA = angle EBA.

Therefore, angle AHB = 180° - angle BAD - angle EBA.

Now, angle BAD = 90° - angle ABD (from triangle ABD), and angle EBA = 90° - angle BAE (from triangle ABE).

But angle ABD is part of angle ABC, which is 52°, and angle BAE is part of angle BAC, which is 54°.

Wait, so angle BAD = 90° - angle ABD, and angle EBA = 90° - angle BAE.

Therefore, angle AHB = 180° - [90° - angle ABD] - [90° - angle BAE] = 180° - 90° + angle ABD - 90° + angle BAE = angle ABD + angle BAE.

But angle ABD + angle BAE = angle ABD + angle BAE. Let me see what that is.

But angle ABD is part of angle ABC = 52°, and angle BAE is part of angle BAC = 54°.

But angle ABD + angle BAE = angle ABD + angle BAE. How can I relate this to the other angles?

Wait, maybe angle ABD + angle BAE = angle ABC + angle BAC - something?

Alternatively, let me consider that angle BAD + angle ABD = 90°, and angle BAE + angle EBA = 90°, but I need to find angle ABD + angle BAE.

Alternatively, maybe I can find angle ABD and angle BAE in terms of other angles.

Wait, angle ABD is the angle between AB and BD, and angle BAE is the angle between BA and AE. Not sure.

Alternatively, let me think about the fact that in triangle ABC, the sum of angles:

angle ABC = angle ABD + angle DBC (but D is on BC, so angle DBC is zero? Wait, no. Wait, D is on BC, so angle ABD is angle between AB and BD, and angle DBC is angle between BD and BC, but since D is on BC, angle DBC is zero? Wait, no, if D is on BC, then BD is part of BC, so angle between BD and BC is zero. Wait, I think I need to think differently.

Alternatively, since angle BAD = 90° - angle ABD, and angle BAE = 90° - angle ABE.

Wait, if I can find angle ABD and angle ABE, then I can find angle BAD and angle BAE.

But angle ABD is part of angle ABC = 52°, and angle ABE is part of angle ABC as well?

Wait, no. Wait, angle ABE is part of angle ABC? Wait, BE is an altitude from B to AC. So E is on AC. Therefore, angle ABE is the angle between AB and BE, which is part of angle ABC? Wait, no. Because BE is going from B to AC, so angle between BE and AB is less than angle ABC.

Alternatively, maybe we can use trigonometry to find angle ABD and angle ABE.

Alternatively, in triangle ABD, right-angled at D:

tan(angle ABD) = AD / BD

Similarly, in triangle ABE, right-angled at E:

tan(angle ABE) = AE / BE

But without knowing the lengths, this might not help. However, maybe using the earlier coordinate calculation, we found angle AHB ≈ 106°, which is 180° - angle C.

Alternatively, another way to think: In triangle ABC, the measure of angle AHB equals 180° minus the measure of angle ACB. Therefore, angle AHB = 180° - 74° = 106°.

Given that both coordinate geometry and the formula give 106°, I can be confident that this is the correct answer.

Therefore, the measure of angle AHB is 106 degrees.

**Final Answer**
\boxed{106}

To solve for $\angle AHB$ in triangle $ABC$, where the altitudes $\overline{AD}$ and $\overline{BE}$ intersect at the orthocenter $H$, we proceed with a combination of geometric principles and logical deduction.

---

### Step 1: Know the Given Angles of Triangle $ABC$

We are given:
- $\angle BAC = 54^\circ$
- $\angle ABC = 52^\circ$

Using the triangle **angle sum property**, we compute:

$$
\angle BAC + \angle ABC + \angle BCA = 180^\circ \\
\Rightarrow \angle ACB = 180^\circ - 54^\circ - 52^\circ = 74^\circ
$$

So, $\angle BCA = \angle C = 74^\circ$.

---

### Step 2: Understand the Role of Altitudes and the Orthocenter

Since $\overline{AD}$ and $\overline{BE}$ are altitudes, they intersect at the **orthocenter** $H$ of triangle $ABC$.

A **key geometric property** relating the orthocenter and the angles of the triangle is:

$$
\angle AHB = 180^\circ - \angle ACB
$$

This relationship comes from examining triangle $AHB$ and the fact that the altitudes intersect at the orthocenter. Additionally, this can be derived using cyclic quadrilateral properties or by considering that the angles formed at the orthocenter are supplementary to the triangle's original angles at the opposite vertex.

---

### Step 3: Apply the Formula

Using the above property:

$$
\angle AHB = 180^\circ - \angle ACB = 180^\circ - 74^\circ = 106^\circ
$$

---

### Final Answer:

$$
\boxed{106}
$$

[Assistant]
Explanation:
Okay, so I have this geometry problem here. Let me try to figure it out step by step. The problem says:

Altitudes AD and BE of triangle ABC intersect at H. If angle BAC is 54 degrees and angle ABC is 52 degrees, then what is angle AHB?

Alright, let me start by drawing triangle ABC. Let me recall that in any triangle, the altitudes intersect at a point called the orthocenter. So here, H is the orthocenter because AD and BE are altitudes intersecting at H.

First, maybe I should sketch triangle ABC with the given angles. Let me note down the given angles:

Angle BAC (which is angle A) is 54 degrees, and angle ABC (which is angle B) is 52 degrees. So angle C can be found since the sum of angles in a triangle is 180 degrees. Let me calculate angle C first.

So angle C = 180° - angle A - angle B = 180° - 54° - 52°. Let me compute that. 54 + 52 is 106, so 180 - 106 is 74. So angle C is 74 degrees. Got that.

Now, the problem is about angle AHB. Let me recall that H is the orthocenter. So angle AHB is the angle between the altitudes AD and BE. Hmm. How can I find this angle?

I need to recall some properties of orthocenters and maybe some relationships between the angles. Let me think. In triangle ABC, the orthocenter H has some relations with the original angles. Maybe I can use the fact that in certain configurations, angles formed at the orthocenter relate to the original angles.

Alternatively, maybe I can find some cyclic quadrilaterals here? Because altitudes create right angles with the sides. Let me think.

Since AD is an altitude from A to BC, that means AD is perpendicular to BC. Similarly, BE is an altitude from B to AC, so BE is perpendicular to AC. Therefore, AD ⊥ BC and BE ⊥ AC.

Now, let me consider quadrilateral formed by H, the feet of the altitudes, and other points. Wait, but maybe I need to look at triangle AHB. Let me focus on triangle AHB. What angles do I know in this triangle?

I need to find angle at H, which is angle AHB. To find this angle, perhaps I can find other angles in triangle AHB. For example, if I can find angles at A and B in triangle AHB, then I can subtract from 180 degrees to get angle AHB.

But wait, in triangle AHB, the angles at A and B are not the same as the original angles of triangle ABC. Let me think. Let me denote some variables. Let me denote angle HAB and angle HBA. Then angle AHB = 180° - angle HAB - angle HBA.

But what are angle HAB and angle HBA?

Since AD is the altitude from A to BC, then AD is perpendicular to BC, so angle ADC is 90°, and angle ADB is 90°. Similarly, BE is perpendicular to AC, so angle BEA is 90°, and angle BEC is 90°.

Let me consider angle HAB. Since AD is an altitude, angle between AD and AB is angle BAD. Wait, angle BAD is part of angle BAC, which is 54 degrees. Let me try to find angle BAD.

Wait, angle BAD is the angle between AB and AD. But since AD is perpendicular to BC, maybe I can find angle BAD using some relations. Let me recall that in triangle ABC, the altitude AD divides the triangle into two right triangles: ABD and ADC.

In triangle ABD, angle ABD is angle ABC, which is 52 degrees. Wait, no. Wait, angle ABD is part of angle ABC. Wait, actually, angle ABD is angle between AB and BD, but BD is part of BC. Hmm. Maybe it's easier to use trigonometry here.

Alternatively, maybe I can use the fact that in triangle ABC, the measure of angle between two altitudes can be related to the original angles. Wait, I remember that in a triangle, the measure of angle AHB is equal to 180° minus angle ACB. Wait, is that correct?

Wait, let me check. Let me recall that in triangle ABC, the orthocenter H creates some angles. For example, in acute triangles, the orthocenter lies inside the triangle. Let me suppose that triangle ABC is acute. Given that angles A and B are 54° and 52°, which are both less than 90°, and angle C is 74°, also less than 90°. So yes, triangle ABC is acute, so H lies inside the triangle.

Now, there's a property that angle AHB = 180° - angle ACB. Wait, if that's the case, then angle AHB would be 180° - 74° = 106°. But let me verify if this is correct. Let me think why that might be.

Alternatively, maybe angle AHB = 180° - angle ABC? Wait, no. Let me think more carefully.

Let me consider quadrilateral formed by H, the feet of the altitudes, and maybe some other points. Wait, let me consider triangles involving H.

Since AD and BE are altitudes, then H lies at their intersection. Let me consider triangle AHB. Let me try to find angles HAB and HBA.

Let me note that angle HAB is equal to angle between AB and altitude AD. Since AD is an altitude, AD is perpendicular to BC. Therefore, angle BAD is equal to 90° minus angle ABC? Wait, no. Wait, angle BAD is part of angle BAC.

Alternatively, maybe I can look at triangle ABD, which is right-angled at D. In triangle ABD, angle at D is 90°, angle at B is angle ABD, which is angle ABC? Wait, no. Wait, angle ABD is part of angle ABC. Wait, angle ABC is angle at B, which is 52°, and since D is on BC, angle ABD is the same as angle ABC? Wait, no. Wait, point D is on BC, so BD is part of BC. Therefore, angle ABD is equal to angle ABC? Wait, no. Wait, angle at B in triangle ABC is angle between AB and BC. But angle ABD is angle between AB and BD. Since BD is part of BC, then angle ABD is the same as angle ABC. Wait, that seems incorrect because BD is just a segment along BC, so angle between AB and BD is the same as angle between AB and BC, which is angle ABC. Wait, is that true?

Wait, suppose I have triangle ABC, with point D on BC. Then angle ABD is the same as angle ABC if D is on BC. Wait, no. Wait, if D is on BC, then BD is just a segment from B to D on BC. So angle between AB and BD is the same as angle between AB and BC, which is angle ABC. Wait, but that seems to suggest angle ABD is equal to angle ABC, but that's only if D is at point C. Wait, no, if D is somewhere along BC, then angle ABD is still part of angle ABC. Wait, maybe I need to think more carefully.

Let me take coordinates. Maybe assigning coordinates to triangle ABC could help. Alternatively, maybe using trigonometric relations.

Alternatively, maybe using the fact that in triangle ABC, angles at H can be found by considering that the orthocenter creates similar triangles or something.

Alternatively, here's a method I remember: in triangle ABC, the measure of angle AHB is equal to 180° minus angle ACB. Wait, let me check with an example. Suppose triangle ABC is equilateral, all angles 60°, then orthocenter is also the centroid and incenter. Then angle AHB would be... Well, in that case, angle AHB would be 120°, since each angle at the orthocenter in an equilateral triangle is 120°, which is 180° - 60°, so yes, 180° - angle C. So if angle C is 60°, angle AHB is 120°, which matches. Similarly, maybe in general angle AHB = 180° - angle ACB?

Wait, if that's the case, then in our problem, angle ACB is 74°, so angle AHB would be 180° - 74° = 106°, which would be the answer. But I need to verify this.

Alternatively, let me try to derive it. Let me think about quadrilateral formed by the orthocenter and some points.

Since AD and BE are altitudes, intersecting at H. So, let me consider quadrilateral AEHD, where E is the foot of BE on AC, and D is the foot of AD on BC. Wait, maybe not. Alternatively, let me consider angles at H.

Since BE is an altitude, BE ⊥ AC. So angle BEA is 90°, and similarly AD ⊥ BC, so angle ADB is 90°. Let me consider triangle AHB. Let me think about angles in triangle AHB.

Alternatively, since BE is perpendicular to AC, then angle between BE and AC is 90°, so angle between BE and AB can be calculated. Wait, angle between BE and AB is angle ABE. Similarly, angle between AD and AB is angle BAD. Then, maybe angle HAB is equal to angle BAD?

Wait, angle BAD is part of angle BAC. Let me think. Let me try to find angle HAB and angle HBA.

Let me denote angle HAB as x and angle HBA as y. Then angle AHB = 180° - x - y.

So if I can find x and y, I can find angle AHB.

Now, what is angle HAB? HAB is the angle between HA and AB. But HA is part of the altitude AD. Therefore, angle HAB is equal to angle between altitude AD and AB. Since AD is perpendicular to BC, maybe angle HAB relates to angle ABC?

Wait, let me recall that in triangle ABD, which is right-angled at D, angle at D is 90°, angle at B is angle ABD, and angle at A is angle BAD.

So in triangle ABD: angle ABD + angle BAD + 90° = 180°, so angle ABD + angle BAD = 90°.

But angle ABD is part of angle ABC. Since ABC is 52°, angle ABD is equal to angle ABC? No, wait. If D is on BC, then angle ABD is the same as angle ABC? Wait, no, unless D is at C. Wait, D is the foot of altitude from A to BC. So unless the triangle is right-angled at B, which it is not. Wait, in triangle ABC, angle at B is 52°, and altitude from A to BC is AD. Therefore, BD is a segment of BC, and angle ABD is part of angle ABC.

Wait, let me denote angle ABD as something. Let me think. Let me consider triangle ABC with altitude AD. Then angle BAD = angle BAC - angle DAC. Wait, but maybe I need another approach.

Alternatively, since in triangle ABD, angle BAD + angle ABD = 90°, so angle BAD = 90° - angle ABD.

Similarly, angle ABD is equal to angle ABC, but no, because D is not necessarily C. Wait, angle ABD is equal to angle ABC only if D coincides with C. But D is the foot of altitude from A to BC. Therefore, BD can be found using trigonometry.

Alternatively, maybe I can use trigonometric ratios. Let me assign some lengths. Let me assume BC is of some length, say, let me assign BC = 1 for simplicity. Wait, but maybe that's too vague. Alternatively, maybe assign AB, BC, AC using the given angles?

Alternatively, let me use the fact that in triangle ABC, the altitude from A to BC is AD. Then, the length of AD can be expressed in terms of the sides and angles, but maybe this is overcomplicating.

Wait, maybe I can find angle BAD. Since angle BAD + angle ABD = 90°, and angle ABD is part of angle ABC. Let me denote angle ABD as θ, so angle BAD = 90° - θ. But angle ABC is 52°, so θ is part of it. Wait, but unless I know something else about θ, I can't determine it.

Alternatively, maybe considering triangle ABE. Since BE is an altitude, it is perpendicular to AC. So angle AEB is 90°. In triangle ABE, angle at E is 90°, angle at B is angle ABE, and angle at A is angle BAE.

Similarly, angle BAE is part of angle BAC which is 54°, so angle BAE = angle BAC - angle EAC? Wait, maybe not. Let me think. If BE is altitude from B to AC, then E is on AC, so angle ABE is part of angle ABC?

Wait, this seems getting too tangled. Maybe there is a better approach.

Wait, let me recall that in any triangle, the measure of angle AHB is equal to 180° minus the measure of angle ACB. Wait, if that is true, then, as I thought earlier, angle AHB = 180° - 74° = 106°, so the answer would be 106 degrees.

But why is angle AHB equal to 180° - angle C?

Let me try to derive this. Let me consider quadrilateral HEBD. Wait, but BE and AD are altitudes intersecting at H. Since BE ⊥ AC and AD ⊥ BC, then HE is part of BE, and HD is part of AD. Therefore, angles at E and D are right angles. Therefore, quadrilateral HEBD has two right angles at E and D. But unless other angles are known, not sure.

Alternatively, consider triangle AHB. Let me find the measure of angle AHB by considering the cyclic quadrilaterals.

Wait, here's a better idea. In triangle ABC, the orthocenter H has the property that angles formed at H relate to the original triangle's angles.

For instance, in triangle ABC, the angles at H can be calculated by:

angle AHB = 180° - angle ACB

angle BHC = 180° - angle BAC

angle AHC = 180° - angle ABC

Is this a valid formula?

Let me check with the equilateral triangle case. If all angles are 60°, then angle AHB should be 180° - 60° = 120°, which is correct. If we take a right-angled triangle, say with angle C = 90°, then angle AHB = 180° - 90° = 90°, but in that case, the orthocenter is at point C, so angle AHB would be angle ACB, which is 90°, so that also matches.

Wait, but in the case of a right-angled triangle, orthocenter is at the right-angled vertex, so angle AHB would be angle ACB. Wait, but if angle C is 90°, then angle AHB is 90°, which is equal to angle ACB. However, according to the formula angle AHB = 180° - angle ACB, that would be 180° - 90° = 90°, which matches. So it seems correct.

Therefore, in general, angle AHB = 180° - angle ACB. Therefore, in our problem, angle AHB = 180° - angle C = 180° - 74° = 106°, so the answer is 106 degrees.

But let me check this formula again because sometimes I might mix up the angles.

Alternatively, here's another way to think about it. In triangle ABC, let's consider the orthocenter H. Then, quadrilateral HECD is cyclic? Wait, maybe not. Alternatively, since AD and BE are altitudes, then angles AHB, let me consider the following.

Since BE is perpendicular to AC, then angle between BE and AC is 90°, and AD is perpendicular to BC, angle between AD and BC is 90°. So, in triangle AHB, angles at H.

Wait, if I can find some relationship between angles. Let me think about the fact that in triangle ABC, the angles formed by the altitudes can be related to the original angles.

Alternatively, using the fact that angle between two altitudes is equal to 180° minus the original angle. Wait, if I consider the angles in triangle AHB:

Let me consider angle HAB. Since AD is an altitude, angle BAD = ?

Similarly, angle HBA = ?

Earlier, I thought of using triangle ABD. In triangle ABD, right-angled at D, angle BAD = 90° - angle ABD.

Similarly, angle ABD is angle between AB and BD, but BD is part of BC. Since angle ABC is 52°, then angle between AB and BC is 52°, so angle ABD is some portion of that? Wait, if we can find angle ABD?

Wait, perhaps using the fact that in triangle ABC, the altitude from A divides BC into segments. The ratio of BD to DC can be found using trigonometry? Maybe.

Let me consider triangle ABC, with angle A = 54°, angle B = 52°, angle C = 74°, and let me denote sides as per standard notation: BC = a, AC = b, AB = c.

By the Law of Sines, a / sin A = b / sin B = c / sin C.

So:

a / sin 54° = b / sin 52° = c / sin 74°.

Let me assign some lengths. Let me assume that the length of BC is a = 1 (for simplicity). Then:

a = 1 = BC

Therefore, from Law of Sines:

sin 54° / 1 = sin 52° / b => b = sin 52° / sin 54°

Similarly, c = sin 74° / sin 54°

Let me compute approximate values to see:

First, sin 54° ≈ 0.8090

sin 52° ≈ 0.7880

sin 74° ≈ 0.9613

Therefore:

b ≈ 0.7880 / 0.8090 ≈ 0.9738

c ≈ 0.9613 / 0.8090 ≈ 1.188

So sides:

BC = a = 1

AC = b ≈ 0.9738

AB = c ≈ 1.188

Now, let me find coordinates of the triangle. Let me place point B at (0,0), point C at (1,0). Then I need to find coordinates of point A.

Given that AB ≈ 1.188 and AC ≈ 0.9738. Let me denote coordinates of A as (x, y). Then distance from A to B is sqrt(x² + y²) ≈ 1.188, and distance from A to C is sqrt((x - 1)² + y²) ≈ 0.9738.

So:

x² + y² ≈ (1.188)² ≈ 1.411

(x - 1)² + y² ≈ (0.9738)² ≈ 0.9483

Subtracting the second equation from the first:

x² + y² - [(x - 1)² + y²] ≈ 1.411 - 0.9483

Expanding:

x² - (x² - 2x + 1) ≈ 0.4627

Simplify:

x² - x² + 2x - 1 ≈ 0.4627

So:

2x - 1 ≈ 0.4627 => 2x ≈ 1.4627 => x ≈ 0.7313

Then, plugging back x into x² + y² ≈ 1.411:

(0.7313)^2 + y² ≈ 1.411

Calculating 0.7313² ≈ 0.5348, so y² ≈ 1.411 - 0.5348 ≈ 0.8762 => y ≈ sqrt(0.8762) ≈ 0.936.

So coordinates of A are approximately (0.7313, 0.936). Let me check distance from A to C:

Distance from A(0.7313, 0.936) to C(1,0):

sqrt((1 - 0.7313)^2 + (0 - 0.936)^2) = sqrt((0.2687)^2 + (-0.936)^2) ≈ sqrt(0.0722 + 0.876) ≈ sqrt(0.948) ≈ 0.973, which matches the earlier value of b ≈ 0.9738. Good.

So coordinates:

B(0,0), C(1,0), A(≈0.7313, ≈0.936)

Now, need to find equations of altitudes AD and BE, then find their intersection H, then compute angle AHB.

Let me find altitude AD first. Altitude from A to BC. Since BC is the base from (0,0) to (1,0), which is horizontal. Therefore, the altitude from A to BC is vertical? Wait, no. Wait, BC is horizontal, so its slope is 0, so the altitude from A to BC is vertical if BC is horizontal? Wait, the altitude is perpendicular to BC. Since BC is horizontal, slope 0, then the altitude is vertical, slope undefined. Therefore, the altitude from A is a vertical line passing through A(0.7313, 0.936). Wait, but if BC is horizontal, then the altitude from A to BC is vertical? Wait, no, if BC is horizontal, then the altitude from A would be vertical only if BC is horizontal and the altitude is vertical. Wait, the altitude from A is perpendicular to BC. Since BC is horizontal, the altitude is vertical. Therefore, its equation is x = 0.7313. Therefore, it meets BC at D(0.7313, 0). Therefore, coordinates of D are (0.7313, 0).

Similarly, find equation of altitude BE. BE is the altitude from B to AC. First, find the equation of AC. Points A(0.7313, 0.936) and C(1, 0). Let me compute the slope of AC:

Slope of AC: (0 - 0.936) / (1 - 0.7313) = (-0.936) / (0.2687) ≈ -3.485.

Therefore, the altitude from B(0,0) to AC is perpendicular to AC, so its slope is the negative reciprocal of -3.485, which is 1/3.485 ≈ 0.287.

Therefore, equation of altitude BE is y = 0.287x.

Now, altitude AD is x = 0.7313. Therefore, intersection point H is at x = 0.7313, and y = 0.287 * 0.7313 ≈ 0.287 * 0.7313 ≈ let me compute that:

0.287 * 0.7 = 0.2009, 0.287 * 0.0313 ≈ approx 0.00896. So total ≈ 0.2009 + 0.00896 ≈ 0.2098. So H is approximately at (0.7313, 0.2098).

Now, we need to find angle AHB. Points A(0.7313, 0.936), H(0.7313, 0.2098), B(0,0).

Wait, let me note coordinates:

Point A: (x_A, y_A) ≈ (0.7313, 0.936)

Point H: (x_H, y_H) ≈ (0.7313, 0.2098)

Point B: (0, 0)

So, angle AHB is the angle at point H between points A and B.

Wait, to compute angle AHB, which is the angle at H between HA and HB.

To calculate this angle, I can use coordinates. Let me compute vectors HA and HB, then use the dot product formula.

Vector HA is A - H = (0.7313 - 0.7313, 0.936 - 0.2098) = (0, 0.7262)

Vector HB is B - H = (0 - 0.7313, 0 - 0.2098) = (-0.7313, -0.2098)

Now, angle between vectors HA and HB can be found by:

cos(theta) = (HA • HB) / (|HA| |HB|)

Compute HA • HB: (0)(-0.7313) + (0.7262)(-0.2098) ≈ 0 - 0.7262 * 0.2098 ≈ -0.1524

Compute |HA|: sqrt(0² + 0.7262²) = 0.7262

Compute |HB|: sqrt((-0.7313)^2 + (-0.2098)^2) ≈ sqrt(0.5348 + 0.044) ≈ sqrt(0.5788) ≈ 0.7608

Therefore,

cos(theta) ≈ (-0.1524) / (0.7262 * 0.7608) ≈ (-0.1524) / (0.5525) ≈ -0.2758

Therefore, theta ≈ arccos(-0.2758) ≈ 106 degrees.

Yes! Because arccos(-0.2758) is approximately 106 degrees. Let me check with calculator:

cos(106°): cos(90° + 16°) = -sin(16°) ≈ -0.2756. Which is very close to -0.2758. Therefore, angle theta is approximately 106°, which matches our previous result of 180° - 74° = 106°. Therefore, this confirms that angle AHB is indeed 106 degrees.

Therefore, the answer is 106 degrees.

But let me check once again if there's another way to get this result without coordinates.

Earlier, I thought that angle AHB = 180° - angle ACB. If that formula is correct, then angle AHB = 180° - 74° = 106°, which matches the coordinate calculation.

But why is that formula true? Let me try to derive it.

In triangle ABC, let H be the orthocenter. Then, consider quadrilateral formed by H and the feet of the altitudes. Alternatively, consider the following:

Let me recall that in triangle ABC, angles at H can be related to the original angles. For example, in triangle HBC, angle HBC is equal to 90° - angle ACB. Wait, no, maybe not. Wait, let me think.

Alternatively, consider that in triangle ABC, since BE is an altitude, angle ABE = 90° - angle BAC. Wait, angle ABE is part of angle ABC. Wait, let me try to think step by step.

In triangle ABE, right-angled at E, angle at E is 90°, so angle ABE = 90° - angle BAE. But angle BAE is part of angle BAC, which is 54°, so angle BAE = angle BAC - angle EAC, but I don't know angle EAC.

Wait, this seems complicated. Let me instead consider cyclic quadrilaterals.

Since AD and BE are altitudes, intersecting at H. Then, since AD ⊥ BC and BE ⊥ AC, then points A, B, H, and the feet of the altitudes lie on certain circles?

Alternatively, consider that in triangle ABC, the orthocenter H creates several right angles. For instance, angle AHB is supplementary to angle ACB.

Wait, here's another approach. In triangle ABC, since AD and BE are altitudes, then angles:

In triangle AHB, angle AHB is equal to 180° minus angle HAB minus angle HBA. As I mentioned earlier.

But angle HAB is equal to angle between HA and AB. Since HA is part of altitude AD, which is perpendicular to BC. So angle between AD and AB is angle BAD.

Similarly, angle HBA is equal to angle between HB and BA, which is angle between HB and BA. Since HB is part of altitude BE, which is perpendicular to AC. Therefore, angle between BE and BA is angle EBA.

Therefore, angle HAB = angle BAD, and angle HBA = angle EBA.

Therefore, angle AHB = 180° - angle BAD - angle EBA.

Now, angle BAD = 90° - angle ABD (from triangle ABD), and angle EBA = 90° - angle BAE (from triangle ABE).

But angle ABD is part of angle ABC, which is 52°, and angle BAE is part of angle BAC, which is 54°.

Wait, so angle BAD = 90° - angle ABD, and angle EBA = 90° - angle BAE.

Therefore, angle AHB = 180° - [90° - angle ABD] - [90° - angle BAE] = 180° - 90° + angle ABD - 90° + angle BAE = angle ABD + angle BAE.

But angle ABD + angle BAE = angle ABD + angle BAE. Let me see what that is.

But angle ABD is part of angle ABC = 52°, and angle BAE is part of angle BAC = 54°.

But angle ABD + angle BAE = angle ABD + angle BAE. How can I relate this to the other angles?

Wait, maybe angle ABD + angle BAE = angle ABC + angle BAC - something?

Alternatively, let me consider that angle BAD + angle ABD = 90°, and angle BAE + angle EBA = 90°, but I need to find angle ABD + angle BAE.

Alternatively, maybe I can find angle ABD and angle BAE in terms of other angles.

Wait, angle ABD is the angle between AB and BD, and angle BAE is the angle between BA and AE. Not sure.

Alternatively, let me think about the fact that in triangle ABC, the sum of angles:

angle ABC = angle ABD + angle DBC (but D is on BC, so angle DBC is zero? Wait, no. Wait, D is on BC, so angle ABD is angle between AB and BD, and angle DBC is angle between BD and BC, but since D is on BC, angle DBC is zero? Wait, no, if D is on BC, then BD is part of BC, so angle between BD and BC is zero. Wait, I think I need to think differently.

Alternatively, since angle BAD = 90° - angle ABD, and angle BAE = 90° - angle ABE.

Wait, if I can find angle ABD and angle ABE, then I can find angle BAD and angle BAE.

But angle ABD is part of angle ABC = 52°, and angle ABE is part of angle ABC as well?

Wait, no. Wait, angle ABE is part of angle ABC? Wait, BE is an altitude from B to AC. So E is on AC. Therefore, angle ABE is the angle between AB and BE, which is part of angle ABC? Wait, no. Because BE is going from B to AC, so angle between BE and AB is less than angle ABC.

Alternatively, maybe we can use trigonometry to find angle ABD and angle ABE.

Alternatively, in triangle ABD, right-angled at D:

tan(angle ABD) = AD / BD

Similarly, in triangle ABE, right-angled at E:

tan(angle ABE) = AE / BE

But without knowing the lengths, this might not help. However, maybe using the earlier coordinate calculation, we found angle AHB ≈ 106°, which is 180° - angle C.

Alternatively, another way to think: In triangle ABC, the measure of angle AHB equals 180° minus the measure of angle ACB. Therefore, angle AHB = 180° - 74° = 106°.

Given that both coordinate geometry and the formula give 106°, I can be confident that this is the correct answer.

Therefore, the measure of angle AHB is 106 degrees.

**Final Answer**
\boxed{106}

To solve for $\angle AHB$ in triangle $ABC$, where the altitudes $\overline{AD}$ and $\overline{BE}$ intersect at the orthocenter $H$, we proceed with a combination of geometric principles and logical deduction.

---

### Step 1: Know the Given Angles of Triangle $ABC$

We are given:
- $\angle BAC = 54^\circ$
- $\angle ABC = 52^\circ$

Using the triangle **angle sum property**, we compute:

$$
\angle BAC + \angle ABC + \angle BCA = 180^\circ \\
\Rightarrow \angle ACB = 180^\circ - 54^\circ - 52^\circ = 74^\circ
$$

So, $\angle BCA = \angle C = 74^\circ$.

---

### Step 2: Understand the Role of Altitudes and the Orthocenter

Since $\overline{AD}$ and $\overline{BE}$ are altitudes, they intersect at the **orthocenter** $H$ of triangle $ABC$.

A **key geometric property** relating the orthocenter and the angles of the triangle is:

$$
\angle AHB = 180^\circ - \angle ACB
$$

This relationship comes from examining triangle $AHB$ and the fact that the altitudes intersect at the orthocenter. Additionally, this can be derived using cyclic quadrilateral properties or by considering that the angles formed at the orthocenter are supplementary to the triangle's original angles at the opposite vertex.

---

### Step 3: Apply the Formula

Using the above property:

$$
\angle AHB = 180^\circ - \angle ACB = 180^\circ - 74^\circ = 106^\circ
$$

---

### Final Answer:

$$
\boxed{106}
$$

Solution:
First, we build a diagram:

[asy]

size(150); defaultpen(linewidth(0.8));

pair B = (0,0), C = (3,0), A = (1.8,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);

draw(A--B--C--cycle);

draw(A--P^^B--Q);

label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,E); label("$H$",H,NW);

draw(rightanglemark(C,P,H,3.5));

draw(rightanglemark(H,Q,C,3.5));

[/asy]

We have $\angle AHB = \angle DHE$, and from quadrilateral $CDHE$, we have  \begin{align*}
\angle DHE &= 360^\circ - \angle HEC - \angle ECD - \angle CDH \\
&= 360^\circ - 90^\circ - \angle ACB - 90^\circ\\
&= 180^\circ - \angle ACB.
\end{align*}From triangle $ABC$, we have $180^\circ - \angle ACB = \angle BAC + \angle ABC = 54^\circ + 52^\circ = \boxed{106^\circ}$.