Okay, let's try to solve this geometry problem. The diagram shows points A, B, C, D with AD=BD=CD, and angle BCA is 40 degrees. We need to find the measure of angle BAC. Hmm, let me visualize the diagram first. There's triangle ABC with points D on BC such that AD=BD=CD. So D is the midpoint of BC, right? Wait, if BD=CD, then D is the midpoint. But also, AD=BD=CD, which means AD=BD=CD. So triangle ABD and triangle ADC are both isosceles with AD=BD and AD=CD respectively. 

Let me sketch this mentally. Point D is the midpoint of BC. Then AD is equal in length to BD and CD. So triangle ABD is isosceles with AB=AD? Wait, no. Wait, AD=BD, so triangle ABD is isosceles with base AB? Wait, no. Wait, in triangle ABD, sides AD and BD are equal, so the base is AB. So the base angles at A and B are equal. Similarly, in triangle ADC, AD=CD, so the base angles at A and C are equal. 

But angle BCA is 40 degrees. That's angle at point C in triangle ABC. So angle at C is 40 degrees. Since D is the midpoint of BC, BD=DC. And AD=BD=DC. So AD is equal to half of BC. Wait, no. If BD=DC, then BC=2*BD. But AD=BD, so AD=BD=DC. So AD=BD=DC= (1/2) BC. So BC is twice the length of AD. 

Hmm. Let's denote some variables. Let me assign variables to the angles to make this clearer. Let's let angle BAC = x degrees. Then, since triangle ABC has angle at C as 40 degrees, the remaining angle at B would be 180 - x - 40 = 140 - x degrees. 

Now, since D is the midpoint of BC, BD=DC. Also, AD=BD=DC. So in triangle ABD, sides AD=BD, so it's an isosceles triangle. Therefore, angles at A and B are equal. Wait, in triangle ABD, the sides AD and BD are equal, so the base angles at A and B are equal. Therefore, angle at A (which is angle BAD) equals angle at B (which is angle ABD). 

Similarly, in triangle ADC, sides AD=DC, so angles at A and C are equal. So angle at A (angle CAD) equals angle at C (angle ACD). 

But angle BAC is the sum of angles BAD and CAD. Let me denote angle BAD as y and angle CAD as z. Then, angle BAC = y + z. 

From triangle ABD, since it's isosceles with AD=BD, angle BAD = angle ABD = y. Similarly, in triangle ADC, angle CAD = angle ACD = z. 

Now, in triangle ABC, angle at B is angle ABC = angle ABD + angle DBC. Wait, but D is the midpoint of BC, so BD=DC. However, in triangle ABD, angle ABD is y, and in triangle ADC, angle ACD is z. But angle at B in triangle ABC is angle ABC = angle ABD + angle DBC. Wait, but D is on BC, so angle DBC is part of angle ABC. Wait, actually, angle at B in triangle ABC is angle ABC, which is angle ABD. Wait, no. Wait, in triangle ABD, angle ABD is y, but angle ABC is the angle at B in triangle ABC, which is the same as angle ABD. Wait, no. Wait, point D is on BC, so angle ABC is the same as angle ABD, because D is on BC. So angle ABC = angle ABD = y. Similarly, in triangle ADC, angle ACD = z, which is part of angle ACB in triangle ABC. Since angle ACB is given as 40 degrees, then angle ACD = z = 40 degrees. Wait, is that right? Wait, angle ACD is part of angle ACB. Since D is on BC, angle ACB is split into angle ACD and angle DCB. Wait, but since AD=CD, triangle ADC is isosceles with AD=CD, so angles at A and C are equal. Therefore, angle at C (which is angle ACD) equals angle at A (angle CAD). So angle CAD = angle ACD = z. But angle ACB is 40 degrees, which is angle ACD + angle DCB. Wait, but angle DCB is part of triangle DBC. Wait, maybe I need to approach this differently.

Wait, maybe it's better to consider triangle ABC first. Let's denote angle BAC as x, angle ABC as y, and angle ACB as 40 degrees. Then, x + y + 40 = 180, so x + y = 140. 

Now, D is the midpoint of BC, so BD=DC. Also, AD=BD=DC. So AD=BD=DC. Therefore, triangle ABD and triangle ADC are both isosceles with AD=BD and AD=DC respectively. 

In triangle ABD, AD=BD, so angles at A and B are equal. So angle BAD = angle ABD. Let's denote angle BAD = angle ABD = a. Then, angle at D in triangle ABD is 180 - 2a. 

Similarly, in triangle ADC, AD=DC, so angles at A and C are equal. Let's denote angle CAD = angle ACD = b. Then, angle at D in triangle ADC is 180 - 2b. 

But angle at D in triangle ABD is the same as angle at D in triangle ADC, right? Because D is a common point. Wait, no. Wait, in triangle ABD, angle at D is angle ADB, and in triangle ADC, angle at D is angle ADC. Since points A, B, D, C are connected with lines, angle ADB and angle ADC are supplementary if they are on a straight line. Wait, but in this case, since D is on BC, points B, D, C are colinear. So angles ADB and ADC are supplementary. Therefore, angle ADB + angle ADC = 180 degrees. 

From triangle ABD: angle ADB = 180 - 2a. From triangle ADC: angle ADC = 180 - 2b. Therefore, (180 - 2a) + (180 - 2b) = 180. Simplifying: 360 - 2a - 2b = 180. Therefore, 2a + 2b = 180. Divide both sides by 2: a + b = 90. 

But in triangle ABC, angle BAC = a + b = x. Therefore, x = 90 degrees. Wait, but that can't be right because angle BAC is part of triangle ABC which already has angle BCA as 40 degrees, so the sum of angles would be x + y + 40 = 180. If x=90, then y=50. But is that possible? Let me check again.

Wait, maybe I made a mistake here. Let's see. If a + b = 90, then angle BAC = x = a + b = 90. Then angle ABC = y = 140 - x = 50. Then angle ACB is 40, so 90 + 50 + 40 = 180, which works. But does this configuration hold with AD=BD=CD?

Wait, if angle BAC is 90 degrees, then triangle ABC would have angles 90, 50, 40. Then BC would be the hypotenuse, but let's see. If AD is the median to BC and also equal to BD and DC, does that hold?

Alternatively, maybe there's a different approach. Let's consider the coordinates. The Asymptote code gives coordinates for the points. Let me parse that. The points are:

A: (8.2635, 9.8481)
B: (0,0)
C: (20,0)
D: (10,0)

So BC is from (0,0) to (20,0), so BD=10, DC=10. Therefore, D is at (10,0). AD is from (10,0) to (8.2635,9.8481). Let's compute the length AD. The distance between (10,0) and (8.2635,9.8481) is sqrt[(10 - 8.2635)^2 + (0 - 9.8481)^2] = sqrt[(1.7365)^2 + (9.8481)^2]. Let me compute that:

1.7365 squared is approximately 3.016, and 9.8481 squared is approximately 96.98. Adding them gives about 100. So sqrt(100)=10. Therefore, AD=10. Similarly, BD is the distance from (0,0) to (10,0), which is 10, and DC is from (10,0) to (20,0), which is 10. So indeed, AD=BD=CD=10. So in the coordinate system, AD=BD=CD=10. 

Given that, we can compute angle BAC. Let's compute the coordinates of A: (8.2635,9.8481). Points B: (0,0), C: (20,0). So angle BAC is the angle at point A between points B and C. Wait, no. Wait, angle BAC is the angle at point A between points B and C? Wait, no. Wait, angle BAC is the angle at point A between BA and CA. Wait, no. Wait, angle BAC is at point A between BA and CA? Wait, no. Wait, angle BAC is at point A between BA and CA? Wait, no, angle BAC is at point A between BA and CA. Wait, no, angle BAC is at point A between BA and CA. Wait, no, angle at A between BA and CA is angle BAC. Wait, but in the Asymptote code, the points are labeled A, B, C, D with A at the top. So angle BAC is the angle at point A between BA and CA. Wait, but point A is connected to B and C. So angle BAC is the angle at point A between BA and CA. Wait, but in the Asymptote code, the diagram is drawn with A at the top, connected to B (0,0), C (20,0), and D (10,0). So angle BAC is the angle at A between BA and CA. Wait, but that's not possible because BA is from A to B, and CA is from A to C. Wait, but in the Asymptote code, the lines are drawn from A to B, A to C, and A to D. So angle BAC is indeed the angle at A between BA and CA. So to compute angle BAC, we can use the coordinates.

Given the coordinates of A (8.2635,9.8481), B (0,0), and C (20,0). Let's compute vectors AB and AC.

Vector AB is from A to B: (0 - 8.2635, 0 - 9.8481) = (-8.2635, -9.8481)

Vector AC is from A to C: (20 - 8.2635, 0 - 9.8481) = (11.7365, -9.8481)

Then, angle BAC is the angle between vectors AB and AC at point A. To find this angle, we can use the dot product formula:

cos(theta) = (AB · AC) / (|AB| |AC|)

First, compute the dot product AB · AC:

(-8.2635)(11.7365) + (-9.8481)(-9.8481)

= (-8.2635)(11.7365) + (9.8481)^2

Let me compute each term:

First term: -8.2635 * 11.7365 ≈ -8.2635 * 11.7365 ≈ Let's compute 8 * 11.7365 = 93.892, 0.2635 * 11.7365 ≈ 3.099. So total ≈ 93.892 + 3.099 ≈ 96.991. So first term ≈ -96.991.

Second term: 9.8481^2 ≈ 96.98 (since 9.8481 is approximately 9.85, and 9.85^2 = 97.0225, but exact value is 96.98)

So total dot product ≈ -96.991 + 96.98 ≈ -0.011. Approximately zero. So cos(theta) ≈ -0.011 / (|AB| |AC|)

Compute |AB|: sqrt((-8.2635)^2 + (-9.8481)^2) = sqrt(68.28 + 96.98) = sqrt(165.26) ≈ 12.86

Compute |AC|: sqrt((11.7365)^2 + (-9.8481)^2) = sqrt(137.73 + 96.98) = sqrt(234.71) ≈ 15.31

So |AB| |AC| ≈ 12.86 * 15.31 ≈ 197.2

So cos(theta) ≈ -0.011 / 197.2 ≈ -0.000056. Therefore, theta ≈ arccos(-0.000056) ≈ 90.0003 degrees. So approximately 90 degrees. 

But in the Asymptote code, the angle BCA is given as 40 degrees. Wait, angle at point C between BC and AC. Let's check that. Point C is at (20,0). AC is from (20,0) to (8.2635,9.8481). BC is from (20,0) to (0,0). So vector BC is (-20,0), and vector AC is (-11.7365,9.8481). The angle at point C is the angle between vectors CB and CC. Wait, point C is connected to B and A. So vectors CB is from C to B: (0 - 20, 0 - 0) = (-20, 0). Vector CA is from C to A: (8.2635 - 20, 9.8481 - 0) = (-11.7365,9.8481). So angle at C is the angle between vectors CB (-20,0) and CA (-11.7365,9.8481). 

Compute the angle between these vectors using the dot product:

cos(theta) = (CB · CA) / (|CB| |CA|)

CB · CA = (-20)(-11.7365) + (0)(9.8481) = 234.73 + 0 = 234.73

|CB| = sqrt((-20)^2 + 0^2) = 20

|CA| = sqrt((-11.7365)^2 + 9.8481^2) ≈ sqrt(137.73 + 96.98) ≈ sqrt(234.71) ≈ 15.31

So cos(theta) ≈ 234.73 / (20 * 15.31) ≈ 234.73 / 306.2 ≈ 0.765

Therefore, theta ≈ arccos(0.765) ≈ 40 degrees. Which matches the given angle BCA = 40 degrees. So that checks out.

Therefore, angle BAC is approximately 90 degrees. But according to my earlier reasoning, if a + b = 90, then angle BAC = 90 degrees. However, in reality, in the Asymptote code, angle BAC is computed as approximately 90.0003 degrees, which is practically 90 degrees. Therefore, the answer is 90 degrees. But let's check again with the given coordinates.

Wait, but angle BAC is the angle at point A. Wait, in the Asymptote code, point A is connected to B, C, and D. So angle BAC is the angle at point A between BA and CA. Wait, but in the coordinates, point A is at (8.2635,9.8481). So vector AB is from A to B: (0 - 8.2635, 0 - 9.8481) = (-8.2635, -9.8481). Vector AC is from A to C: (20 - 8.2635, 0 - 9.8481) = (11.7365, -9.8481). The angle between these two vectors is indeed approximately 90 degrees, as we calculated. So angle BAC is 90 degrees. 

But let's confirm with another approach. Since AD=BD=CD=10, and coordinates of D are (10,0). So AD is the line from (10,0) to (8.2635,9.8481), which has length 10. Then, in triangle ABD, which is isosceles with AD=BD=10, so triangle ABD has sides AB, BD=10, AD=10. Similarly, triangle ADC has sides AC, CD=10, AD=10. 

If angle BAC is 90 degrees, then triangle ABC would be a right-angled triangle at A. But in the given coordinates, point A is at (8.2635,9.8481), and the coordinates of B and C are (0,0) and (20,0). So the triangle is not right-angled at A. Wait, but according to the coordinates, angle BAC is approximately 90 degrees. Wait, but according to the coordinates, the vectors AB and AC have a dot product of approximately zero, which suggests that angle BAC is 90 degrees. However, the triangle isn't right-angled at A because the coordinates don't satisfy the Pythagorean theorem. Wait, let's check.

Compute AB length: sqrt(8.2635^2 + 9.8481^2) ≈ sqrt(68.28 + 96.98) ≈ sqrt(165.26) ≈ 12.86

Compute AC length: sqrt(11.7365^2 + 9.8481^2) ≈ sqrt(137.73 + 96.98) ≈ sqrt(234.71) ≈ 15.31

Compute BC length: 20 units.

If angle BAC is 90 degrees, then by Pythagoras, AB^2 + AC^2 should equal BC^2. Let's check:

AB^2 + AC^2 ≈ 165.26 + 234.71 ≈ 399.97

BC^2 = 20^2 = 400

So 399.97 ≈ 400, which is very close. Therefore, angle BAC is indeed 90 degrees.

Therefore, the measure of angle BAC is 90 degrees. 

But let's see if this makes sense with the given information. If angle BAC is 90 degrees, then triangle ABC is right-angled at A, so angle ABC + angle ACB = 90 degrees. Given that angle ACB is 40 degrees, angle ABC would be 50 degrees. 

But in the diagram, D is the midpoint of BC, and AD=BD=CD=10. So in the right-angled triangle at A, the median to the hypotenuse is equal to half the hypotenuse. Wait, yes! In a right-angled triangle, the median from the right angle to the hypotenuse is equal to half the hypotenuse. So in triangle ABC, right-angled at A, the median AD to hypotenuse BC should be equal to BC/2. Since BC is 20, BC/2 is 10, which matches AD=10. Therefore, this is consistent. 

Therefore, angle BAC is indeed 90 degrees. 

But let me verify once again with the given coordinates. The coordinates of A are (8.2635,9.8481). Let's check if triangle ABC is right-angled at A.

Compute AB^2: (8.2635)^2 + (9.8481)^2 ≈ 68.28 + 96.98 ≈ 165.26

Compute AC^2: (11.7365)^2 + (9.8481)^2 ≈ 137.73 + 96.98 ≈ 234.71

Compute BC^2: 20^2 = 400

Check if AB^2 + AC^2 ≈ BC^2: 165.26 + 234.71 ≈ 399.97 ≈ 400. Yes, so it's approximately a right-angled triangle at A. Therefore, angle BAC is 90 degrees. 

Therefore, the measure of angle BAC is 90 degrees.

**Final Answer**
\boxed{90^\circ}
Therefore, the measure of $\angle BAC$ is $\boxed{90^\circ}$.
</think>