<Question>
Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$.  If $\angle BAC = 54^\circ$ and $\angle ABC = 52^\circ$, then what is $\angle AHB$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the measure of angle AHB in triangle ABC where AD and BE are the altitudes intersecting at H. Given angles BAC is 54 degrees and ABC is 52 degrees. Hmm, let me start by visualizing the triangle.

<prompt_1.1> Visualize triangle ABC with the given angles and identify the altitudes AD and BE. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, triangle ABC. Let me label the vertices A, B, and C. Angle at A is 54 degrees, angle at B is 52 degrees. Since the sum of angles in a triangle is 180, angle at C should be 180 - 54 - 52. Let me calculate that: 54 + 52 = 106, so 180 - 106 = 74 degrees. So angle ACB is 74 degrees.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, altitudes AD and BE intersect at H. Altitudes are perpendicular to the opposite sides, right? So AD is perpendicular to BC, and BE is perpendicular to AC. H is the orthocenter of the triangle since that's where the altitudes meet.

<prompt_2.1> Confirm that H is the orthocenter of triangle ABC and recall its properties. </prompt_2.1>

But how does that help me find angle AHB? Let me think. Maybe I can find some relationships between the angles in the triangle and the angles formed at the orthocenter.

Alternatively, perhaps I can consider the quadrilaterals formed by the altitudes. For example, since AD and BE are altitudes, the quadrilateral formed by A, H, the foot of the altitude from A, and the foot of the altitude from B might have some properties.
Wait, but I need to be careful here. Let me try a different approach.

Let me recall that in a triangle, the orthocenter creates several smaller triangles, each similar to the original triangle. Maybe triangle AHB is similar to triangle ACB or something like that. Wait, not sure. Let me check.

<prompt_2.2> Explore the possibility of triangle AHB being similar to triangle ACB. </prompt_2.2>

Alternatively, maybe I can compute angle AHB by considering the sum of angles around point H. But I need more information about the other angles at H.

Alternatively, perhaps I can use the fact that in the orthocenter configuration, the angles at H relate to the original angles of the triangle.

<prompt_2.3> Consider using the properties of the orthocenter to relate angles at H to the original triangle's angles. </prompt_2.3>
-----

Alternatively, maybe I can compute angle AHB by considering triangle AHB. To find angle AHB, we can use the fact that in triangle AHB, the sum of angles is 180 degrees. So if we can find the other two angles at A and B, then we can subtract their sum from 180 to get angle AHB.

<prompt_2.4> Calculate angle AHB by considering triangle AHB and using the sum of angles in a triangle. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Right, so H is the orthocenter. But I need to find angle AHB. Hmm. Let me recall that the orthocenter is the intersection point of the altitudes. So, H is inside the triangle since the triangle is acute? Wait, are all angles acute here? Let's check. Angles at A, B, and C are 54, 52, and 74 degrees respectively. All are less than 90, so yes, the triangle is acute, so the orthocenter H is inside the triangle.
</execution_2.1>

<execution_2.2>
Another idea: maybe consider the triangle formed by the orthocenter. For instance, in triangle ABC, the orthocenter H creates triangle AHB. The angles at H might relate to the original angles of ABC. Let me think.

Alternatively, maybe I can use the fact that angles at H can be related to the original angles of the triangle through cyclic quadrilaterals. For example, since AD and BE are altitudes, the quadrilateral CDHE (if E is the foot on AC and D is the foot on BC) might be cyclic? Wait, not sure.
</execution_2.2>

<execution_2.3>
Wait, angle AHB is equal to 180 degrees minus angle ACB. Is that a general property? Let me think. If H is the orthocenter, then angle AHB should be equal to 180 degrees minus angle ACB. Because the lines AH and BH are the altitudes from A and B, so they form right angles with BC and AC respectively. Therefore, the angle between AH and BH would be supplementary to angle ACB.
</execution_2.3>

<execution_2.4>
Wait, but angle at A in triangle AHB is not the same as angle BAC. Wait, no. Point H is inside the triangle (since the triangle is acute, as all angles are less than 90 degrees). So angle at A in triangle AHB is the same as angle BAH. Similarly, angle at B in triangle AHB is angle ABH.

But how do we find angles BAH and ABH?
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, no. Wait, angle at H. Let me think again. Let's see.

<prompt_3.1> Use the properties of cyclic quadrilaterals to find angle AHB. </prompt_3.1>

Based on execution_2.3:
Wait, is that true? Let me verify.

<prompt_3.2> Verify the property that angle AHB is equal to 180 degrees minus angle ACB. </prompt_3.2>
-----

Based on execution_2.4:
Let me see.

Since AD is the altitude from A to BC, angle BAD is 90 degrees. Similarly, BE is the altitude from B to AC, so angle ABE is 90 degrees.

<prompt_3.3> Find angles BAH and ABH in triangle AHB. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
Since AD is perpendicular to BC, angle ADB is 90 degrees. Similarly, BE is perpendicular to AC, so angle BEA is 90 degrees. Now, quadrilateral CDHE (if H is the orthocenter) would have two right angles at D and E. Wait, but I need to check if that's cyclic. For a quadrilateral to be cyclic, the sum of opposite angles must be 180 degrees. If angles at D and E are 90 degrees, then unless the other two angles also sum to 180, it might not be cyclic. Hmm, maybe that's not the way to go.
</execution_3.1>

<execution_3.2>
Let me recall that in the orthocenter configuration, the angles at H are related to the original angles. Specifically, angle AHB + angle ACB = 180 degrees. Because the two angles subtended by the same arc or something? Wait, maybe it's because quadrilateral HACB is cyclic? Wait, no. Wait, if we consider the circle with diameter AH, but maybe that's not helpful.

Alternatively, let's consider the triangle. Since AD and BE are altitudes, then angles AHB and ACB are related. Let me try to find angle AHB.

Let me consider the triangle AHB. The angles at A and B in this triangle are not the original angles of triangle ABC, but rather parts of them. Wait, angle at A in triangle AHB is angle BAH, and angle at B is angle ABH. Then angle AHB is 180 - (angle BAH + angle ABH). So if I can find angle BAH and angle ABH, then I can find angle AHB.
</execution_3.2>

<execution_3.3>
But in triangle ABC, angle BAC is 54 degrees. So angle BAC is 54 degrees, which is angle at A. If angle BAD is 90 degrees, but wait, angle BAD is not 90 degrees. Wait, AD is the altitude from A to BC, so angle ADC and angle ADB are 90 degrees. Wait, angle BAD is part of angle BAC. Let me correct that.

So, angle BAC is 54 degrees. The altitude from A to BC is AD, so angle ADC is 90 degrees. Therefore, angle BAD is adjacent to angle DAC. Wait, but angle BAC is 54 degrees, so angle BAD + angle DAC = 54 degrees. Similarly, angle ABC is 52 degrees, so the altitude BE from B to AC makes angle BEA 90 degrees. Therefore, angle ABE is part of angle ABC. Wait, angle ABC is 52 degrees, so angle ABE + angle EBC = 52 degrees. But since BE is the altitude, angle BEC is 90 degrees.

Wait, perhaps I need to find angles BAH and ABH. Let me denote angle BAH as x and angle ABH as y. Then, angle AHB is 180 - x - y. But how to find x and y?

Alternatively, maybe there's a better way. Let's recall that in the orthocenter configuration, the angles at H are related to the original angles of the triangle. Specifically, angle AHB is equal to 180 degrees minus angle ACB. Let me check if that's a known property.

Yes, actually, in a triangle, the measure of the angle formed at the orthocenter by two altitudes is equal to 180 degrees minus the measure of the original angle at the third vertex. So, angle AHB = 180 - angle ACB. Since angle ACB is 74 degrees, angle AHB would be 180 - 74 = 106 degrees. Therefore, the answer is 106 degrees.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Let me try to find angle BAH. Since AD is the altitude, triangle ABD is right-angled at D. But angle BAC is 54 degrees. Wait, angle BAC is 54, which is angle at A. If I can find angle BAD, then angle BAH would be part of that.

<prompt_4.1> Find angle BAH using the properties of right triangles and the given angles in triangle ABC. </prompt_4.1>

Based on execution_3.3:
Wait, but let me verify this. Let me consider another approach to confirm.

<prompt_4.2> Verify the result using another approach to confirm the measure of angle AHB. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Wait, maybe not. Let me think again.

Alternatively, since BE is the altitude from B to AC, then angle BEA is 90 degrees. So in triangle BEA, angle at E is 90 degrees, angle at B is 52 degrees, so angle at A in that triangle (angle EAB) would be 180 - 90 - 52 = 38 degrees. But angle EAB is the same as angle BAC, which is 54 degrees? Wait, that can't be. Wait, no. Wait, angle EAB is part of angle BAC. Wait, no.

Wait, BE is the altitude from B to AC, so E is the foot of the altitude on AC. Therefore, BE is perpendicular to AC, so angle BEA is 90 degrees. Therefore, triangle BEA is right-angled at E. So in triangle BEA, angle at B is 52 degrees, angle at E is 90 degrees, so angle at A (angle EAB) is 180 - 90 - 52 = 38 degrees. Therefore, angle EAB is 38 degrees. But angle BAC is given as 54 degrees. Therefore, angle between BA and BE is 38 degrees, which is part of angle BAC. Wait, but angle BAC is the angle at A between BA and CA. Since E is on AC, then angle EAB is part of angle BAC. Therefore, angle BAC = angle EAB + angle EAC. Wait, but angle EAB is 38 degrees, and angle BAC is 54 degrees, so angle EAC would be 54 - 38 = 16 degrees. Hmm, interesting.

But how does that help me with angle BAH?
</execution_4.1>

<execution_4.2>
Let's construct the orthocenter H. Then, angles at H: angle AHB, angle BHC, angle CHA. There's a property that angle AHB + angle BHC + angle CHA = 360 degrees. But that might not help directly. Alternatively, since H is the orthocenter, each angle at H is equal to 180 degrees minus the corresponding angle at the vertex opposite. So angle AHB = 180 - angle ACB, angle BHC = 180 - angle BAC, and angle AHC = 180 - angle ABC. Therefore, angle AHB = 180 - 74 = 106 degrees.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, maybe not. Let me think again.

Wait, angle EAB is 38 degrees, which is the angle between BA and BE. Since BE is the altitude, which is from B to AC, so BE is perpendicular to AC. Therefore, angle between BE and AC is 90 degrees. Therefore, angle between BA and AC is angle BAC, which is 54 degrees. So angle between BA and BE is 38 degrees, so angle between BE and AC is 54 - 38 = 16 degrees.

<prompt_5.1> Calculate angle EAC using the information from the right triangle BEA. </prompt_5.1>
-----

Based on execution_4.2:
Therefore, angle AHB = 180 - 74 = 106 degrees.
So, \boxed{106} degrees.

But let me just verify once more.

<prompt_5.2> Verify the result using another approach to ensure the measure of angle AHB is correct. </prompt_5.2>
</Plan_5>

<Execution_5>
<execution_5.1>
But angle BEA is 90 degrees, so angle EAB is 38 degrees, so angle between BE and BA is 38 degrees, and angle between BE and AC is 16 degrees. Hmm, not sure if that's helpful yet.

Alternatively, maybe I can find angle ABH. Since BE is the altitude, angle EBC is 52 degrees (angle ABC). Wait, angle EBC is part of angle ABC. Since BE is the altitude, angle EBC is adjacent to angle EBA. Wait, angle EBC + angle EBA = angle ABC = 52 degrees. But in triangle BEA, angle at E is 90 degrees, angle at B is 52 degrees, so angle at A is 38 degrees. Wait, angle at A in triangle BEA is angle EAB = 38 degrees, which is part of angle BAC = 54 degrees. Therefore, angle between BA and BE is 38 degrees, so angle between BE and BC is 54 - 38 = 16 degrees. Therefore, angle EBC = 16 degrees.
</execution_5.1>

<execution_5.2>
Alternatively, using the fact that in triangle ABC, the orthocenter H creates triangle AHB. The angles at H: angle AHB. There's a formula that angle AHB = 180 - angle ACB. Since angle ACB is 74, so 180 - 74 = 106. Yes. So that's consistent.

Alternatively, another way to see this is that in the circumcircle of triangle ABC, the angle at the center is twice the angle at the circumference, but maybe that's complicating. Alternatively, since H is the orthocenter, the angles at H are related to the original angles. So, angle AHB is equal to 180 - angle ACB. Therefore, 180 - 74 = 106 degrees. Therefore, the answer is 106 degrees.
</execution_5.2>
</Execution_5>

<Plan_6>
Based on execution_5.2:
**Final Answer**
\boxed{106}
</Plan_6>

<Execution_6>
No prompt to execute.
</Execution_6>

<Final_answer>
Therefore, the measure of $\angle AHB$ is $\boxed{106}$ degrees.
</Final_answer>
</think>