Explanation:
Okay, so I need to find the area of triangle ACD in this trapezoid ABCD where AD is parallel to BC. They also told me that AC is perpendicular to CD, AC bisects angle BAD, and the area of the trapezoid is 42. Hmm, let me try to visualize this first.

First, let me sketch the trapezoid. Since AD is parallel to BC, and it's a trapezoid, so the bases are AD and BC. Now, AC is a diagonal that's perpendicular to CD. Also, AC bisects angle BAD. So angle BAC is equal to angle CAD. Let me note that down.

Let me label the trapezoid. Let me think: points A, B, C, D. AD is parallel to BC. So maybe A and D are the two bottom vertices, and B and C are the top ones? Wait, but trapezoid can be oriented in different ways. Maybe I should assign coordinates to make it easier? Coordinates might help here.

Alternatively, maybe using properties of trapezoids and angle bisectors. Since AC is an angle bisector of angle BAD, and it's also perpendicular to CD. Let me think about that.

Let me recall that in a trapezoid with bases AD and BC, the legs are AB and CD. Wait, but if AD is parallel to BC, then AB and CD are the non-parallel sides. But they mentioned that AC is perpendicular to CD. So AC is perpendicular to CD, meaning that triangle ACD is a right triangle? Because AC is perpendicular to CD, so angle ACD is 90 degrees. So triangle ACD is right-angled at C. Therefore, the area of triangle ACD would be (AC * CD)/2. But I don't know AC or CD yet. But maybe I can find relations between them?

Also, AC bisects angle BAD. Let me denote angle BAD as 2θ, so that angle BAC = angle CAD = θ. Since AC is an angle bisector, maybe I can use the angle bisector theorem? The angle bisector theorem relates the lengths of the sides of a triangle to the segments created by the bisector on the opposite side. But here, AC is the angle bisector of angle BAD, so maybe in triangle ABD? Wait, but AC is a diagonal of the trapezoid. Let me think.

Wait, in triangle ABD, if AC is the angle bisector of angle BAD, but AC is not necessarily intersecting BD. Wait, maybe I need to consider triangle ABD? Wait, but AC is a diagonal of the trapezoid, so it connects A to C. So maybe triangle ABC? Hmm, not sure. Maybe I need to think about coordinates.

Let me try coordinates. Let me place point A at the origin (0, 0). Since AD is parallel to BC, and AC is perpendicular to CD. Let me assume AD is horizontal? Wait, but if AC is perpendicular to CD, then CD is vertical? Maybe. Let me try to assign coordinates step by step.

Let me set coordinate system with point A at (0, 0). Let me assume AD is along the x-axis. So point D is at (d, 0) for some d > 0. Since AD is parallel to BC, BC is also horizontal. Let me denote coordinates of B as (p, h) and C as (q, h), since BC is parallel to AD (so same y-coordinate). Wait, but then AB connects A(0,0) to B(p, h), and CD connects C(q, h) to D(d, 0). Also, AC is perpendicular to CD. Let me write equations for that.

First, vector AC is from A(0,0) to C(q, h), so direction vector (q, h). Vector CD is from C(q, h) to D(d, 0), direction vector (d - q, -h). Since AC is perpendicular to CD, their dot product is zero:

(q)(d - q) + (h)(-h) = 0

So q(d - q) - h² = 0 => qd - q² - h² = 0. Let me note that equation (1): qd - q² - h² = 0.

Also, AC bisects angle BAD. Let me think about angle BAD. Since AD is along the x-axis from A(0,0) to D(d, 0), and AB goes from A(0,0) to B(p, h). So angle BAD is the angle between AD (along x-axis) and AB. The angle bisector AC divides this angle into two equal parts. So the direction vector of AC is (q, h), which should be the angle bisector between AD (direction vector (1, 0)) and AB (direction vector (p, h)).

There is a formula for angle bisector in terms of vectors. Alternatively, maybe using the angle bisector theorem. In triangle ABD, if AC is the angle bisector of angle BAD, then the ratio of AB to AD is equal to the ratio of BC to CD? Wait, no, angle bisector theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. Wait, but here AC is not intersecting BD, it's intersecting BC? Wait, no. Wait, AC connects A to C, which is a vertex of the trapezoid. So maybe in triangle ABD, but point C is not on BD. Maybe I need to think differently.

Alternatively, since AC is the angle bisector of angle BAD, then the direction vector of AC should satisfy the angle bisector condition between vectors AB and AD. Let me recall that if a vector is an angle bisector between two vectors, then it's proportional to the sum of the unit vectors in those directions.

So, vector AD is (d, 0), so unit vector is (1, 0). Vector AB is (p, h), so unit vector is (p / |AB|, h / |AB|). Then the angle bisector direction vector should be (1, 0) + (p / |AB|, h / |AB|). But AC is in direction (q, h). Therefore, (q, h) is parallel to (1 + p / |AB|, 0 + h / |AB|). Therefore, there exists some scalar k such that:

q = k(1 + p / |AB|)

h = k(h / |AB|)

From the second equation: h = k(h / |AB|). Assuming h ≠ 0 (otherwise, the trapezoid would be degenerate), we can divide both sides by h:

1 = k / |AB| => k = |AB|.

Therefore, substituting back into the first equation:

q = |AB| * (1 + p / |AB|) = |AB| + p.

But |AB| is sqrt(p² + h²). Therefore:

q = sqrt(p² + h²) + p.

Hmm, that's an equation relating p, q, h. Let me note that.

So equation (2): q = sqrt(p² + h²) + p.

Also, from equation (1): qd - q² - h² = 0.

So maybe I can use equation (2) to substitute into equation (1). Let me try that.

First, let me write equation (1):

qd - q² - h² = 0 => qd = q² + h².

From equation (2), q = sqrt(p² + h²) + p. Let me denote sqrt(p² + h²) as |AB|, which is the length of AB. Let me denote AB = sqrt(p² + h²) = let's say, m. Then q = m + p.

So equation (1) becomes:

(m + p) * d = (m + p)^2 + h².

Expanding the right-hand side:

(m + p)^2 + h² = m² + 2mp + p² + h².

But m² = p² + h², so substituting:

= (p² + h²) + 2mp + p² + h² = 2p² + 2h² + 2mp.

Therefore, equation (1) becomes:

(m + p) * d = 2p² + 2h² + 2mp.

Divide both sides by 2:

[(m + p) * d]/2 = p² + h² + mp.

But m² = p² + h², so:

[(m + p) * d]/2 = m² + mp.

Let me write that:

(m + p) * d / 2 = m² + m p.

Let me factor the right-hand side:

m² + m p = m(m + p).

Therefore:

(m + p) * d / 2 = m(m + p).

Assuming m + p ≠ 0 (which it isn't, since m and p are lengths and coordinates), we can divide both sides by (m + p):

d / 2 = m => d = 2m.

So d = 2m. But m is |AB| = sqrt(p² + h²). Therefore, d = 2 sqrt(p² + h²).

So that's another relation. So d = 2 sqrt(p² + h²). Let me note that.

So now, we have:

From equation (2): q = m + p = sqrt(p² + h²) + p.

From equation (1): qd = q² + h². But since d = 2m, and q = m + p, let me check if that holds.

Wait, we already derived that from equation (1) and (2), so maybe we can proceed.

Now, we need more relations. Also, since ABCD is a trapezoid with AD || BC. So BC is parallel to AD. Since AD is horizontal (from (0,0) to (d, 0)), then BC must also be horizontal. Therefore, points B(p, h) and C(q, h) have the same y-coordinate h, which we already considered.

Now, since ABCD is a trapezoid, the sides AB and CD are the legs. Wait, but in a trapezoid, legs are the non-parallel sides. Since AD || BC, then AB and CD are the legs. However, they mentioned that AC is perpendicular to CD. So CD is a leg, and AC is perpendicular to CD. So CD is a leg, and AC is a diagonal that is perpendicular to CD.

So, vector CD is (d - q, -h). Earlier, we had that AC is perpendicular to CD, which gave us equation (1). Which we already used.

Now, we need to find the area of the trapezoid ABCD, which is given as 42. The area of a trapezoid is (base1 + base2)/2 * height. Here, the bases are AD and BC. The length of AD is d, and the length of BC is q - p, since BC goes from (p, h) to (q, h). The height of the trapezoid is the vertical distance between the two bases, which is h. Therefore, area is [(d + (q - p))/2] * h = 42.

So let me write that equation:

[(d + q - p)/2] * h = 42.

So equation (3): (d + q - p) * h = 84.

Now, we have several relations:

From earlier:

d = 2m = 2 sqrt(p² + h²).

q = m + p = sqrt(p² + h²) + p.

So let me substitute d and q in terms of p and h into equation (3):

First, compute d + q - p:

d + q - p = 2 sqrt(p² + h²) + [sqrt(p² + h²) + p] - p = 2 sqrt(p² + h²) + sqrt(p² + h²) + p - p = 3 sqrt(p² + h²).

Therefore, equation (3) becomes:

[3 sqrt(p² + h²)] * h = 84.

So 3 h sqrt(p² + h²) = 84 => h sqrt(p² + h²) = 28.

Let me denote sqrt(p² + h²) as m again. Then h * m = 28. But m = sqrt(p² + h²). Therefore, h * sqrt(p² + h²) = 28. Let me square both sides to eliminate the square root:

h² (p² + h²) = 28² = 784.

So equation (4): h² (p² + h²) = 784.

Now, we need another equation to relate p and h. Wait, do we have more relations?

Earlier, we had from equation (1) and others, but maybe we need to use the fact that ABCD is a trapezoid. Wait, but we already used that BC is parallel to AD. Also, we have coordinates for all points. Maybe we need to find relations from the fact that ABCD is a trapezoid? Wait, but we already considered that. Maybe there's another condition?

Wait, perhaps we need to use the fact that AC is perpendicular to CD. Wait, we already used that to get equation (1). So maybe we need another relation? Wait, but we have equation (4): h² (p² + h²) = 784. But we need another equation to solve for p and h. Wait, but maybe we can find another relation from the coordinates?

Wait, perhaps we can find coordinates of point C? Wait, point C is (q, h) where q = m + p = sqrt(p² + h²) + p. Also, since AD is from (0,0) to (d, 0) where d = 2m. So coordinates of D are (2m, 0). Coordinates of C are (sqrt(p² + h²) + p, h). Also, since BC is from (p, h) to (q, h), which is horizontal, as we had.

But maybe there's another condition? Wait, maybe the fact that ABCD is a trapezoid? Wait, but we already used that AD || BC. Maybe we need to ensure that AB and CD are not parallel? Wait, but in a trapezoid, only one pair of sides is parallel. Wait, but AD || BC, so AB and CD are the legs, which are not necessarily parallel. However, since AC is perpendicular to CD, and AC is a diagonal, maybe there's something else?

Alternatively, maybe we can find the area of triangle ACD and relate it to the trapezoid area?

Wait, the problem asks for [ACD], which is the area of triangle ACD. Since triangle ACD is right-angled at C, as AC ⊥ CD. Therefore, area of triangle ACD is (AC * CD)/2.

We can compute AC and CD.

First, AC is from A(0,0) to C(q, h). So length AC is sqrt(q² + h²). But q = m + p, and m = sqrt(p² + h²). Therefore, AC = sqrt( (m + p)^2 + h² ). Let me compute that:

(m + p)^2 + h² = m² + 2mp + p² + h². But m² = p² + h², so substituting:

= (p² + h²) + 2mp + p² + h² = 2p² + 2h² + 2mp = 2(p² + h² + mp). Wait, but earlier we had that d = 2m, and from equation (3) we had h * m = 28. Maybe not helpful yet.

Alternatively, CD is from C(q, h) to D(d, 0). So length CD is sqrt( (d - q)^2 + (0 - h)^2 ) = sqrt( (d - q)^2 + h² ). From equation (1), we had that qd - q² - h² = 0 => qd = q² + h² => d - q = (q² + h²)/q - q = (q² + h² - q²)/q = h² / q. Therefore, d - q = h² / q. Therefore, CD = sqrt( (h² / q)^2 + h² ) = sqrt( h^4 / q² + h² ) = h * sqrt( h² / q² + 1 ) = h * sqrt( (h² + q²)/q² ) = h * sqrt( h² + q² ) / q.

But AC is sqrt(q² + h²). Therefore, CD = h * AC / q. Therefore, area of triangle ACD is (AC * CD)/2 = (AC * (h * AC / q )) / 2 = (h * AC² ) / (2q).

But AC² is q² + h². Therefore, area of ACD is h(q² + h²)/(2q).

But from equation (1): qd = q² + h². Therefore, q² + h² = qd. Therefore, area of ACD is h * qd / (2q) = h d / 2.

So area of triangle ACD is (h d)/2. Wait, that's interesting. So [ACD] = (h d)/2. But the area of the trapezoid is [(AD + BC)/2] * h = [(d + (q - p))/2] * h = 42. Earlier, we found that d + q - p = 3m, and since m = sqrt(p² + h²), but we also had that h * m = 28. Wait, but maybe we can relate [ACD] to the trapezoid area?

Wait, if [ACD] = (h d)/2, and the trapezoid area is [(d + (q - p))/2] * h = 42. From earlier, we had that d + q - p = 3m, and m = sqrt(p² + h²). Also, h * m = 28. So let me write:

Trapezoid area: [ (d + q - p)/2 ] * h = (3m / 2) * h = (3/2) * m * h = 42. But m * h = 28, so (3/2)*28 = 42. Which checks out! Because 3/2 * 28 = 42. So that's consistent. Therefore, the trapezoid area is indeed 42, as given.

But how does that help me find [ACD]? Since [ACD] = (h d)/2. So if I can find h*d, then I can compute [ACD].

But from equation (4): h² (p² + h²) = 784. Also, we have d = 2m = 2 sqrt(p² + h²). Let me denote m = sqrt(p² + h²). Then d = 2m, and equation (4) becomes h² * m² = 784. Therefore, (h m)^2 = 784 => h m = 28, which we already had. So h m = 28. Therefore, h * d = h * 2m = 2 * h m = 2 * 28 = 56. Therefore, [ACD] = (h d)/2 = 56 / 2 = 28. Therefore, the area of triangle ACD is 28?

Wait, but the trapezoid area is 42, and triangle ACD is 28? Let me check if that makes sense. The trapezoid area is 42, and triangle ACD is part of it. Let me think about the other parts.

The trapezoid can be divided into triangles ABC and ACD? Wait, no. If you draw diagonal AC, then the trapezoid is divided into triangle ABC and triangle ACD. Wait, but is that correct? Let me see. In trapezoid ABCD, with bases AD and BC, drawing diagonal AC divides it into triangle ABC and triangle ACD. Therefore, [ABCD] = [ABC] + [ACD] = 42. If [ACD] is 28, then [ABC] would be 14. Is that possible?

Alternatively, maybe there's another way to check. Let me think. If [ACD] = 28, then since [ACD] = (h d)/2, and [ABCD] = [(AD + BC)/2] * h = 42. Let me compute AD = d, BC = q - p. Earlier, we had that d + q - p = 3m. Therefore, AD + BC = d + (q - p) = 3m. Therefore, [(AD + BC)/2] * h = (3m / 2) * h = 42. But since h * m = 28, then (3/2)*28 = 42, which is correct. Therefore, the calculation seems consistent.

But let me check if [ACD] = 28 is correct. Because if [ACD] = (h d)/2, and h d = 56, then yes. But how do we know h d = 56? Because h m = 28, and d = 2m, so h d = h * 2m = 2 * 28 = 56. Therefore, yes. Therefore, [ACD] = 56 / 2 = 28. Therefore, the answer is 28? But let me check again.

Wait, but let me think again. If the trapezoid area is 42, and triangle ACD is 28, then triangle ABC is 14. Is there a way to verify this?

Alternatively, maybe I made a mistake in assuming that [ACD] = (h d)/2. Let me rederive that.

Earlier, I said that CD = h * AC / q. Then area of ACD is (AC * CD)/2 = (AC * h * AC / q)/2 = h * AC² / (2q). Then since AC² = q² + h², and from equation (1): qd = q² + h², so AC² = qd. Therefore, area of ACD is h * qd / (2q) = h d / 2. Yes, that's correct. Therefore, [ACD] = (h d)/2. And since h d = 56, then [ACD] = 28. Therefore, the answer is 28. But let me check if that's possible.

Alternatively, maybe I can find numerical values for h and d. Let me try.

We have h * m = 28, where m = sqrt(p² + h²). Also, d = 2m. So if I can find h and m, then I can compute h*d.

But we have equation (4): h² * m² = 784. But h * m = 28, so (h * m)^2 = 784, which is exactly equation (4). Therefore, that equation is redundant. Therefore, we have infinitely many solutions depending on p and h, but the value of h*d is fixed as 56. Therefore, regardless of the specific values of p and h, [ACD] is always 28. Therefore, the answer is 28.

But let me check with an example. Let me choose specific values for p and h that satisfy the equations.

Let me assume some value for h and m. Let me suppose h = 7, then m = 28 / h = 28 / 7 = 4. Therefore, m = 4. Then, since m = sqrt(p² + h²) = sqrt(p² + 49) = 4. Therefore, sqrt(p² + 49) = 4 => p² + 49 = 16 => p² = -33. Which is impossible. Therefore, h cannot be 7. So my assumption is wrong.

Wait, so h * m = 28, and m = sqrt(p² + h²). Therefore, h * sqrt(p² + h²) = 28. Let me pick h = 4. Then sqrt(p² + 16) = 28 / 4 = 7. Therefore, sqrt(p² + 16) = 7 => p² + 16 = 49 => p² = 33 => p = sqrt(33). So that's possible. Then m = 7, d = 2m = 14. Then h*d = 4*14 = 56. Therefore, [ACD] = 56 / 2 = 28. So that works. Let me check if with these values, the other conditions hold.

So h = 4, p = sqrt(33), m = 7, d = 14. Then q = m + p = 7 + sqrt(33). Then BC length is q - p = 7 + sqrt(33) - sqrt(33) = 7. Therefore, the trapezoid has bases AD = 14 and BC = 7, height h = 4. Therefore, area is (14 + 7)/2 * 4 = 21/2 * 4 = 42. Correct. Also, check if AC is perpendicular to CD.

Coordinates:

A(0, 0), D(14, 0), B(sqrt(33), 4), C(7 + sqrt(33), 4).

Vector AC is from A(0,0) to C(7 + sqrt(33), 4): direction vector (7 + sqrt(33), 4).

Vector CD is from C(7 + sqrt(33), 4) to D(14, 0): direction vector (14 - (7 + sqrt(33)), 0 - 4) = (7 - sqrt(33), -4).

Dot product of AC and CD:

(7 + sqrt(33))(7 - sqrt(33)) + (4)(-4) = [49 - (sqrt(33))²] - 16 = [49 - 33] - 16 = 16 - 16 = 0. Perfect! So they are perpendicular. Also, check if AC bisects angle BAD.

Vector AB is from A(0,0) to B(sqrt(33), 4): direction vector (sqrt(33), 4).

Vector AD is from A(0,0) to D(14, 0): direction vector (14, 0).

We need to check if AC bisects the angle between AB and AD.

The angle bisector direction should be proportional to the sum of the unit vectors of AB and AD.

Unit vector of AD is (1, 0).

Unit vector of AB is (sqrt(33)/sqrt( (sqrt(33))² + 4² ), 4 / sqrt( (sqrt(33))² + 4² )) = (sqrt(33)/sqrt(33 + 16), 4 / sqrt(49)) = (sqrt(33)/7, 4/7).

Therefore, sum of unit vectors is (1 + sqrt(33)/7, 0 + 4/7) = ( (7 + sqrt(33))/7, 4/7 ). Therefore, direction vector is proportional to (7 + sqrt(33), 4), which is exactly the direction vector of AC. Therefore, AC is indeed the angle bisector. Perfect! So all conditions are satisfied.

Therefore, in this case, [ACD] = 28, which is exactly half of h*d = 56. Therefore, the answer is 28. Therefore, I think that is correct.

But let me check another example to be safe. Let me choose h = 14, then m = 28 / 14 = 2. Then m = sqrt(p² + h²) => 2 = sqrt(p² + 196) => p² + 196 = 4 => p² = -192. Not possible. So h cannot be 14. Let me choose h = 2, then m = 28 / 2 = 14. Then sqrt(p² + 4) = 14 => p² = 196 - 4 = 192 => p = sqrt(192) = 8*sqrt(3). Then d = 2m = 28. Then h*d = 2*28 = 56. Therefore, [ACD] = 56 / 2 = 28. Let me check if the trapezoid area is 42.

AD = d = 28, BC = q - p. q = m + p = 14 + 8√3. Therefore, BC = q - p = 14 + 8√3 - 8√3 = 14. Therefore, trapezoid area is (28 + 14)/2 * 2 = (42/2)*2 = 42. Correct. Also check AC perpendicular to CD.

Coordinates:

A(0,0), D(28, 0), B(8√3, 2), C(14 + 8√3, 2).

Vector AC: (14 + 8√3, 2).

Vector CD: D - C = (28 - (14 + 8√3), 0 - 2) = (14 - 8√3, -2).

Dot product: (14 + 8√3)(14 - 8√3) + (2)(-2) = [14² - (8√3)^2] - 4 = [196 - 192] - 4 = 4 - 4 = 0. Correct. Also, angle bisector check.

Vector AB: (8√3, 2). Unit vector: (8√3 / sqrt( (8√3)^2 + 2^2 ), 2 / sqrt( ... )) = (8√3 / sqrt(192 + 4), 2 / sqrt(196)) = (8√3 / 14, 2 / 14) = (4√3 / 7, 1/7).

Unit vector of AD is (1, 0). Sum of unit vectors: (1 + 4√3 / 7, 0 + 1/7) = ( (7 + 4√3)/7, 1/7 ). Direction vector proportional to (7 + 4√3, 1). But AC direction vector is (14 + 8√3, 2) which is 2*(7 + 4√3, 1). Therefore, same direction. Therefore, AC is angle bisector. All conditions satisfied. Therefore, [ACD] = 28. So regardless of the specific values, the area is 28. Therefore, the answer is 28.

But wait, the problem says "compute [triangle ACD]". So 28? But let me check once more. The trapezoid area is 42, and triangle ACD is 28, which is 2/3 of the trapezoid area? Wait, 28 is 2/3 of 42? 42*(2/3) = 28. Yes. But why?

Alternatively, since the trapezoid is divided by AC into triangle ABC and ACD. If [ACD] = 28, then [ABC] = 14. Maybe there's a ratio here. Since AC is an angle bisector, maybe the areas are in some ratio?

Alternatively, since AC bisects angle BAD, and in triangle ABD, but I think we already checked with coordinates. Anyway, since multiple examples confirm that [ACD] = 28, and the derivation seems correct, I think the answer is 28.

But let me check once more with another approach. Let me think about ratios.

Since AC is the angle bisector of angle BAD, and in triangle ABD, but AC meets BD? Wait, no, AC meets BC at point C. Wait, maybe using the angle bisector theorem in triangle ABD? Wait, but point C is not on BD. Maybe not helpful.

Alternatively, since AC is the angle bisector, and AC ⊥ CD, maybe there are some similar triangles?

In triangle ACD, right-angled at C, and angle at A is θ, which is half of angle BAD. Also, in triangle ABC?

Alternatively, since AC bisects angle BAD, and AC ⊥ CD, maybe triangle ACD is related to triangle ABC?

Alternatively, since we found that [ACD] = (h d)/2 and [ABCD] = ( (d + BC)/2 ) * h = 42, and we found that [ACD] = 28, which is 2/3 of the trapezoid area. But why?

Alternatively, since BC = q - p = 7 in the first example, and AD = 14, so BC is half of AD. Therefore, the trapezoid has bases in ratio 2:1. Then area is average of bases times height. If BC is half of AD, then average is (AD + BC)/2 = (14 + 7)/2 = 10.5, times height 4 is 42. Then [ACD] is (h d)/2 = 4*14 / 2 = 28. Which is 2/3 of 42. But why?

Alternatively, since AC divides the trapezoid into two parts, and because of the angle bisector and the perpendicularity, the areas have a specific ratio. But since we have multiple confirmations, I think 28 is correct.

Therefore, I think the answer is \boxed{28}.

**Final Answer**
\boxed{28}

To solve the problem, we are given a trapezoid $ABCD$ with the following properties:

- $AD \parallel BC$
- $\overline{AC} \perp \overline{CD}$
- $\overline{AC}$ bisects $\angle BAD$
- The area of the trapezoid is $[ABCD] = 42$
- We are to find the area of triangle $\triangle ACD$

---

### Step 1: Understand the Geometry

Let’s place the trapezoid in the coordinate plane for clarity:

- Let $A = (0, 0)$
- Since $AD \parallel BC$, and we can assume $AD$ is horizontal, let $D = (d, 0)$
- Let $B = (p, h)$ and $C = (q, h)$, so that $BC$ is also horizontal (same y-coordinate as $B$ and $C$)

This setup ensures that $AD$ and $BC$ are the two bases of the trapezoid, and $AB$ and $CD$ are the legs.

---

### Step 2: Use the Given Conditions

#### (a) $\overline{AC} \perp \overline{CD}$

This implies that the vectors $\vec{AC}$ and $\vec{CD}$ are perpendicular. Using coordinates:

- $\vec{AC} = (q, h)$
- $\vec{CD} = (d - q, -h)$

Their dot product is:

$$
q(d - q) + h(-h) = 0 \Rightarrow qd - q^2 - h^2 = 0 \tag{1}
$$

#### (b) $\overline{AC}$ bisects $\angle BAD$

This implies that the direction vector of $\vec{AC}$ lies between $\vec{AB}$ and $\vec{AD}$. Using vector angle bisector properties, we can derive that:

$$
q = \sqrt{p^2 + h^2} + p \tag{2}
$$

Let’s define $m = \sqrt{p^2 + h^2}$, the length of $\vec{AB}$. Then:

$$
q = m + p \tag{3}
$$

From equation (1), we can also express:

$$
qd = q^2 + h^2 \Rightarrow d = \frac{q^2 + h^2}{q} \tag{4}
$$

Substituting (3) into (4):

$$
d = \frac{(m + p)^2 + h^2}{m + p}
$$

But since $m^2 = p^2 + h^2$, we can simplify this expression further.

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### Step 3: Use the Area of the Trapezoid

The area of trapezoid $ABCD$ is:

$$
[ABCD] = \frac{1}{2}(AD + BC) \cdot h = \frac{1}{2}(d + (q - p)) \cdot h = 42
$$

From earlier, we found:

$$
d + q - p = 3m \Rightarrow \frac{1}{2}(3m) \cdot h = 42 \Rightarrow \frac{3}{2}mh = 42 \Rightarrow mh = 28 \tag{5}
$$

---

### Step 4: Compute the Area of Triangle $ACD$

Since $\overline{AC} \perp \overline{CD}$, triangle $ACD$ is a right triangle with right angle at $C$. Therefore, its area is:

$$
[\triangle ACD] = \frac{1}{2} \cdot AC \cdot CD
$$

We can express both $AC$ and $CD$ in terms of $m$ and $h$:

- $AC = \sqrt{q^2 + h^2} = \sqrt{(m + p)^2 + h^2} = \sqrt{m^2 + 2mp + p^2 + h^2} = \sqrt{2m^2 + 2mp} = \sqrt{2m(m + p)}$
- From earlier, $CD = \frac{h^2}{q}$, and since $q = m + p$, we get:

$$
CD = \frac{h^2}{m + p}
$$

So the area becomes:

$$
[\triangle ACD] = \frac{1}{2} \cdot \sqrt{2m(m + p)} \cdot \frac{h^2}{m + p} = \frac{1}{2} \cdot \frac{h^2 \sqrt{2m(m + p)}}{m + p}
$$

But this is complicated. Instead, we use a more elegant approach.

From earlier, we found:

$$
[\triangle ACD] = \frac{1}{2} \cdot h \cdot d
$$

And from equation (5), $mh = 28$, and from earlier, $d = 2m$, so:

$$
h \cdot d = h \cdot 2m = 2 \cdot mh = 2 \cdot 28 = 56
$$

Thus:

$$
[\triangle ACD] = \frac{1}{2} \cdot 56 = \boxed{28}
$$

---

### Final Answer

$$
\boxed{28}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle ACD in this trapezoid ABCD where AD is parallel to BC. They also told me that AC is perpendicular to CD, AC bisects angle BAD, and the area of the trapezoid is 42. Hmm, let me try to visualize this first.

First, let me sketch the trapezoid. Since AD is parallel to BC, and it's a trapezoid, so the bases are AD and BC. Now, AC is a diagonal that's perpendicular to CD. Also, AC bisects angle BAD. So angle BAC is equal to angle CAD. Let me note that down.

Let me label the trapezoid. Let me think: points A, B, C, D. AD is parallel to BC. So maybe A and D are the two bottom vertices, and B and C are the top ones? Wait, but trapezoid can be oriented in different ways. Maybe I should assign coordinates to make it easier? Coordinates might help here.

Alternatively, maybe using properties of trapezoids and angle bisectors. Since AC is an angle bisector of angle BAD, and it's also perpendicular to CD. Let me think about that.

Let me recall that in a trapezoid with bases AD and BC, the legs are AB and CD. Wait, but if AD is parallel to BC, then AB and CD are the non-parallel sides. But they mentioned that AC is perpendicular to CD. So AC is perpendicular to CD, meaning that triangle ACD is a right triangle? Because AC is perpendicular to CD, so angle ACD is 90 degrees. So triangle ACD is right-angled at C. Therefore, the area of triangle ACD would be (AC * CD)/2. But I don't know AC or CD yet. But maybe I can find relations between them?

Also, AC bisects angle BAD. Let me denote angle BAD as 2θ, so that angle BAC = angle CAD = θ. Since AC is an angle bisector, maybe I can use the angle bisector theorem? The angle bisector theorem relates the lengths of the sides of a triangle to the segments created by the bisector on the opposite side. But here, AC is the angle bisector of angle BAD, so maybe in triangle ABD? Wait, but AC is a diagonal of the trapezoid. Let me think.

Wait, in triangle ABD, if AC is the angle bisector of angle BAD, but AC is not necessarily intersecting BD. Wait, maybe I need to consider triangle ABD? Wait, but AC is a diagonal of the trapezoid, so it connects A to C. So maybe triangle ABC? Hmm, not sure. Maybe I need to think about coordinates.

Let me try coordinates. Let me place point A at the origin (0, 0). Since AD is parallel to BC, and AC is perpendicular to CD. Let me assume AD is horizontal? Wait, but if AC is perpendicular to CD, then CD is vertical? Maybe. Let me try to assign coordinates step by step.

Let me set coordinate system with point A at (0, 0). Let me assume AD is along the x-axis. So point D is at (d, 0) for some d > 0. Since AD is parallel to BC, BC is also horizontal. Let me denote coordinates of B as (p, h) and C as (q, h), since BC is parallel to AD (so same y-coordinate). Wait, but then AB connects A(0,0) to B(p, h), and CD connects C(q, h) to D(d, 0). Also, AC is perpendicular to CD. Let me write equations for that.

First, vector AC is from A(0,0) to C(q, h), so direction vector (q, h). Vector CD is from C(q, h) to D(d, 0), direction vector (d - q, -h). Since AC is perpendicular to CD, their dot product is zero:

(q)(d - q) + (h)(-h) = 0

So q(d - q) - h² = 0 => qd - q² - h² = 0. Let me note that equation (1): qd - q² - h² = 0.

Also, AC bisects angle BAD. Let me think about angle BAD. Since AD is along the x-axis from A(0,0) to D(d, 0), and AB goes from A(0,0) to B(p, h). So angle BAD is the angle between AD (along x-axis) and AB. The angle bisector AC divides this angle into two equal parts. So the direction vector of AC is (q, h), which should be the angle bisector between AD (direction vector (1, 0)) and AB (direction vector (p, h)).

There is a formula for angle bisector in terms of vectors. Alternatively, maybe using the angle bisector theorem. In triangle ABD, if AC is the angle bisector of angle BAD, then the ratio of AB to AD is equal to the ratio of BC to CD? Wait, no, angle bisector theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. Wait, but here AC is not intersecting BD, it's intersecting BC? Wait, no. Wait, AC connects A to C, which is a vertex of the trapezoid. So maybe in triangle ABD, but point C is not on BD. Maybe I need to think differently.

Alternatively, since AC is the angle bisector of angle BAD, then the direction vector of AC should satisfy the angle bisector condition between vectors AB and AD. Let me recall that if a vector is an angle bisector between two vectors, then it's proportional to the sum of the unit vectors in those directions.

So, vector AD is (d, 0), so unit vector is (1, 0). Vector AB is (p, h), so unit vector is (p / |AB|, h / |AB|). Then the angle bisector direction vector should be (1, 0) + (p / |AB|, h / |AB|). But AC is in direction (q, h). Therefore, (q, h) is parallel to (1 + p / |AB|, 0 + h / |AB|). Therefore, there exists some scalar k such that:

q = k(1 + p / |AB|)

h = k(h / |AB|)

From the second equation: h = k(h / |AB|). Assuming h ≠ 0 (otherwise, the trapezoid would be degenerate), we can divide both sides by h:

1 = k / |AB| => k = |AB|.

Therefore, substituting back into the first equation:

q = |AB| * (1 + p / |AB|) = |AB| + p.

But |AB| is sqrt(p² + h²). Therefore:

q = sqrt(p² + h²) + p.

Hmm, that's an equation relating p, q, h. Let me note that.

So equation (2): q = sqrt(p² + h²) + p.

Also, from equation (1): qd - q² - h² = 0.

So maybe I can use equation (2) to substitute into equation (1). Let me try that.

First, let me write equation (1):

qd - q² - h² = 0 => qd = q² + h².

From equation (2), q = sqrt(p² + h²) + p. Let me denote sqrt(p² + h²) as |AB|, which is the length of AB. Let me denote AB = sqrt(p² + h²) = let's say, m. Then q = m + p.

So equation (1) becomes:

(m + p) * d = (m + p)^2 + h².

Expanding the right-hand side:

(m + p)^2 + h² = m² + 2mp + p² + h².

But m² = p² + h², so substituting:

= (p² + h²) + 2mp + p² + h² = 2p² + 2h² + 2mp.

Therefore, equation (1) becomes:

(m + p) * d = 2p² + 2h² + 2mp.

Divide both sides by 2:

[(m + p) * d]/2 = p² + h² + mp.

But m² = p² + h², so:

[(m + p) * d]/2 = m² + mp.

Let me write that:

(m + p) * d / 2 = m² + m p.

Let me factor the right-hand side:

m² + m p = m(m + p).

Therefore:

(m + p) * d / 2 = m(m + p).

Assuming m + p ≠ 0 (which it isn't, since m and p are lengths and coordinates), we can divide both sides by (m + p):

d / 2 = m => d = 2m.

So d = 2m. But m is |AB| = sqrt(p² + h²). Therefore, d = 2 sqrt(p² + h²).

So that's another relation. So d = 2 sqrt(p² + h²). Let me note that.

So now, we have:

From equation (2): q = m + p = sqrt(p² + h²) + p.

From equation (1): qd = q² + h². But since d = 2m, and q = m + p, let me check if that holds.

Wait, we already derived that from equation (1) and (2), so maybe we can proceed.

Now, we need more relations. Also, since ABCD is a trapezoid with AD || BC. So BC is parallel to AD. Since AD is horizontal (from (0,0) to (d, 0)), then BC must also be horizontal. Therefore, points B(p, h) and C(q, h) have the same y-coordinate h, which we already considered.

Now, since ABCD is a trapezoid, the sides AB and CD are the legs. Wait, but in a trapezoid, legs are the non-parallel sides. Since AD || BC, then AB and CD are the legs. However, they mentioned that AC is perpendicular to CD. So CD is a leg, and AC is perpendicular to CD. So CD is a leg, and AC is a diagonal that is perpendicular to CD.

So, vector CD is (d - q, -h). Earlier, we had that AC is perpendicular to CD, which gave us equation (1). Which we already used.

Now, we need to find the area of the trapezoid ABCD, which is given as 42. The area of a trapezoid is (base1 + base2)/2 * height. Here, the bases are AD and BC. The length of AD is d, and the length of BC is q - p, since BC goes from (p, h) to (q, h). The height of the trapezoid is the vertical distance between the two bases, which is h. Therefore, area is [(d + (q - p))/2] * h = 42.

So let me write that equation:

[(d + q - p)/2] * h = 42.

So equation (3): (d + q - p) * h = 84.

Now, we have several relations:

From earlier:

d = 2m = 2 sqrt(p² + h²).

q = m + p = sqrt(p² + h²) + p.

So let me substitute d and q in terms of p and h into equation (3):

First, compute d + q - p:

d + q - p = 2 sqrt(p² + h²) + [sqrt(p² + h²) + p] - p = 2 sqrt(p² + h²) + sqrt(p² + h²) + p - p = 3 sqrt(p² + h²).

Therefore, equation (3) becomes:

[3 sqrt(p² + h²)] * h = 84.

So 3 h sqrt(p² + h²) = 84 => h sqrt(p² + h²) = 28.

Let me denote sqrt(p² + h²) as m again. Then h * m = 28. But m = sqrt(p² + h²). Therefore, h * sqrt(p² + h²) = 28. Let me square both sides to eliminate the square root:

h² (p² + h²) = 28² = 784.

So equation (4): h² (p² + h²) = 784.

Now, we need another equation to relate p and h. Wait, do we have more relations?

Earlier, we had from equation (1) and others, but maybe we need to use the fact that ABCD is a trapezoid. Wait, but we already used that BC is parallel to AD. Also, we have coordinates for all points. Maybe we need to find relations from the fact that ABCD is a trapezoid? Wait, but we already considered that. Maybe there's another condition?

Wait, perhaps we need to use the fact that AC is perpendicular to CD. Wait, we already used that to get equation (1). So maybe we need another relation? Wait, but we have equation (4): h² (p² + h²) = 784. But we need another equation to solve for p and h. Wait, but maybe we can find another relation from the coordinates?

Wait, perhaps we can find coordinates of point C? Wait, point C is (q, h) where q = m + p = sqrt(p² + h²) + p. Also, since AD is from (0,0) to (d, 0) where d = 2m. So coordinates of D are (2m, 0). Coordinates of C are (sqrt(p² + h²) + p, h). Also, since BC is from (p, h) to (q, h), which is horizontal, as we had.

But maybe there's another condition? Wait, maybe the fact that ABCD is a trapezoid? Wait, but we already used that AD || BC. Maybe we need to ensure that AB and CD are not parallel? Wait, but in a trapezoid, only one pair of sides is parallel. Wait, but AD || BC, so AB and CD are the legs, which are not necessarily parallel. However, since AC is perpendicular to CD, and AC is a diagonal, maybe there's something else?

Alternatively, maybe we can find the area of triangle ACD and relate it to the trapezoid area?

Wait, the problem asks for [ACD], which is the area of triangle ACD. Since triangle ACD is right-angled at C, as AC ⊥ CD. Therefore, area of triangle ACD is (AC * CD)/2.

We can compute AC and CD.

First, AC is from A(0,0) to C(q, h). So length AC is sqrt(q² + h²). But q = m + p, and m = sqrt(p² + h²). Therefore, AC = sqrt( (m + p)^2 + h² ). Let me compute that:

(m + p)^2 + h² = m² + 2mp + p² + h². But m² = p² + h², so substituting:

= (p² + h²) + 2mp + p² + h² = 2p² + 2h² + 2mp = 2(p² + h² + mp). Wait, but earlier we had that d = 2m, and from equation (3) we had h * m = 28. Maybe not helpful yet.

Alternatively, CD is from C(q, h) to D(d, 0). So length CD is sqrt( (d - q)^2 + (0 - h)^2 ) = sqrt( (d - q)^2 + h² ). From equation (1), we had that qd - q² - h² = 0 => qd = q² + h² => d - q = (q² + h²)/q - q = (q² + h² - q²)/q = h² / q. Therefore, d - q = h² / q. Therefore, CD = sqrt( (h² / q)^2 + h² ) = sqrt( h^4 / q² + h² ) = h * sqrt( h² / q² + 1 ) = h * sqrt( (h² + q²)/q² ) = h * sqrt( h² + q² ) / q.

But AC is sqrt(q² + h²). Therefore, CD = h * AC / q. Therefore, area of triangle ACD is (AC * CD)/2 = (AC * (h * AC / q )) / 2 = (h * AC² ) / (2q).

But AC² is q² + h². Therefore, area of ACD is h(q² + h²)/(2q).

But from equation (1): qd = q² + h². Therefore, q² + h² = qd. Therefore, area of ACD is h * qd / (2q) = h d / 2.

So area of triangle ACD is (h d)/2. Wait, that's interesting. So [ACD] = (h d)/2. But the area of the trapezoid is [(AD + BC)/2] * h = [(d + (q - p))/2] * h = 42. Earlier, we found that d + q - p = 3m, and since m = sqrt(p² + h²), but we also had that h * m = 28. Wait, but maybe we can relate [ACD] to the trapezoid area?

Wait, if [ACD] = (h d)/2, and the trapezoid area is [(d + (q - p))/2] * h = 42. From earlier, we had that d + q - p = 3m, and m = sqrt(p² + h²). Also, h * m = 28. So let me write:

Trapezoid area: [ (d + q - p)/2 ] * h = (3m / 2) * h = (3/2) * m * h = 42. But m * h = 28, so (3/2)*28 = 42. Which checks out! Because 3/2 * 28 = 42. So that's consistent. Therefore, the trapezoid area is indeed 42, as given.

But how does that help me find [ACD]? Since [ACD] = (h d)/2. So if I can find h*d, then I can compute [ACD].

But from equation (4): h² (p² + h²) = 784. Also, we have d = 2m = 2 sqrt(p² + h²). Let me denote m = sqrt(p² + h²). Then d = 2m, and equation (4) becomes h² * m² = 784. Therefore, (h m)^2 = 784 => h m = 28, which we already had. So h m = 28. Therefore, h * d = h * 2m = 2 * h m = 2 * 28 = 56. Therefore, [ACD] = (h d)/2 = 56 / 2 = 28. Therefore, the area of triangle ACD is 28?

Wait, but the trapezoid area is 42, and triangle ACD is 28? Let me check if that makes sense. The trapezoid area is 42, and triangle ACD is part of it. Let me think about the other parts.

The trapezoid can be divided into triangles ABC and ACD? Wait, no. If you draw diagonal AC, then the trapezoid is divided into triangle ABC and triangle ACD. Wait, but is that correct? Let me see. In trapezoid ABCD, with bases AD and BC, drawing diagonal AC divides it into triangle ABC and triangle ACD. Therefore, [ABCD] = [ABC] + [ACD] = 42. If [ACD] is 28, then [ABC] would be 14. Is that possible?

Alternatively, maybe there's another way to check. Let me think. If [ACD] = 28, then since [ACD] = (h d)/2, and [ABCD] = [(AD + BC)/2] * h = 42. Let me compute AD = d, BC = q - p. Earlier, we had that d + q - p = 3m. Therefore, AD + BC = d + (q - p) = 3m. Therefore, [(AD + BC)/2] * h = (3m / 2) * h = 42. But since h * m = 28, then (3/2)*28 = 42, which is correct. Therefore, the calculation seems consistent.

But let me check if [ACD] = 28 is correct. Because if [ACD] = (h d)/2, and h d = 56, then yes. But how do we know h d = 56? Because h m = 28, and d = 2m, so h d = h * 2m = 2 * 28 = 56. Therefore, yes. Therefore, [ACD] = 56 / 2 = 28. Therefore, the answer is 28? But let me check again.

Wait, but let me think again. If the trapezoid area is 42, and triangle ACD is 28, then triangle ABC is 14. Is there a way to verify this?

Alternatively, maybe I made a mistake in assuming that [ACD] = (h d)/2. Let me rederive that.

Earlier, I said that CD = h * AC / q. Then area of ACD is (AC * CD)/2 = (AC * h * AC / q)/2 = h * AC² / (2q). Then since AC² = q² + h², and from equation (1): qd = q² + h², so AC² = qd. Therefore, area of ACD is h * qd / (2q) = h d / 2. Yes, that's correct. Therefore, [ACD] = (h d)/2. And since h d = 56, then [ACD] = 28. Therefore, the answer is 28. But let me check if that's possible.

Alternatively, maybe I can find numerical values for h and d. Let me try.

We have h * m = 28, where m = sqrt(p² + h²). Also, d = 2m. So if I can find h and m, then I can compute h*d.

But we have equation (4): h² * m² = 784. But h * m = 28, so (h * m)^2 = 784, which is exactly equation (4). Therefore, that equation is redundant. Therefore, we have infinitely many solutions depending on p and h, but the value of h*d is fixed as 56. Therefore, regardless of the specific values of p and h, [ACD] is always 28. Therefore, the answer is 28.

But let me check with an example. Let me choose specific values for p and h that satisfy the equations.

Let me assume some value for h and m. Let me suppose h = 7, then m = 28 / h = 28 / 7 = 4. Therefore, m = 4. Then, since m = sqrt(p² + h²) = sqrt(p² + 49) = 4. Therefore, sqrt(p² + 49) = 4 => p² + 49 = 16 => p² = -33. Which is impossible. Therefore, h cannot be 7. So my assumption is wrong.

Wait, so h * m = 28, and m = sqrt(p² + h²). Therefore, h * sqrt(p² + h²) = 28. Let me pick h = 4. Then sqrt(p² + 16) = 28 / 4 = 7. Therefore, sqrt(p² + 16) = 7 => p² + 16 = 49 => p² = 33 => p = sqrt(33). So that's possible. Then m = 7, d = 2m = 14. Then h*d = 4*14 = 56. Therefore, [ACD] = 56 / 2 = 28. So that works. Let me check if with these values, the other conditions hold.

So h = 4, p = sqrt(33), m = 7, d = 14. Then q = m + p = 7 + sqrt(33). Then BC length is q - p = 7 + sqrt(33) - sqrt(33) = 7. Therefore, the trapezoid has bases AD = 14 and BC = 7, height h = 4. Therefore, area is (14 + 7)/2 * 4 = 21/2 * 4 = 42. Correct. Also, check if AC is perpendicular to CD.

Coordinates:

A(0, 0), D(14, 0), B(sqrt(33), 4), C(7 + sqrt(33), 4).

Vector AC is from A(0,0) to C(7 + sqrt(33), 4): direction vector (7 + sqrt(33), 4).

Vector CD is from C(7 + sqrt(33), 4) to D(14, 0): direction vector (14 - (7 + sqrt(33)), 0 - 4) = (7 - sqrt(33), -4).

Dot product of AC and CD:

(7 + sqrt(33))(7 - sqrt(33)) + (4)(-4) = [49 - (sqrt(33))²] - 16 = [49 - 33] - 16 = 16 - 16 = 0. Perfect! So they are perpendicular. Also, check if AC bisects angle BAD.

Vector AB is from A(0,0) to B(sqrt(33), 4): direction vector (sqrt(33), 4).

Vector AD is from A(0,0) to D(14, 0): direction vector (14, 0).

We need to check if AC bisects the angle between AB and AD.

The angle bisector direction should be proportional to the sum of the unit vectors of AB and AD.

Unit vector of AD is (1, 0).

Unit vector of AB is (sqrt(33)/sqrt( (sqrt(33))² + 4² ), 4 / sqrt( (sqrt(33))² + 4² )) = (sqrt(33)/sqrt(33 + 16), 4 / sqrt(49)) = (sqrt(33)/7, 4/7).

Therefore, sum of unit vectors is (1 + sqrt(33)/7, 0 + 4/7) = ( (7 + sqrt(33))/7, 4/7 ). Therefore, direction vector is proportional to (7 + sqrt(33), 4), which is exactly the direction vector of AC. Therefore, AC is indeed the angle bisector. Perfect! So all conditions are satisfied.

Therefore, in this case, [ACD] = 28, which is exactly half of h*d = 56. Therefore, the answer is 28. Therefore, I think that is correct.

But let me check another example to be safe. Let me choose h = 14, then m = 28 / 14 = 2. Then m = sqrt(p² + h²) => 2 = sqrt(p² + 196) => p² + 196 = 4 => p² = -192. Not possible. So h cannot be 14. Let me choose h = 2, then m = 28 / 2 = 14. Then sqrt(p² + 4) = 14 => p² = 196 - 4 = 192 => p = sqrt(192) = 8*sqrt(3). Then d = 2m = 28. Then h*d = 2*28 = 56. Therefore, [ACD] = 56 / 2 = 28. Let me check if the trapezoid area is 42.

AD = d = 28, BC = q - p. q = m + p = 14 + 8√3. Therefore, BC = q - p = 14 + 8√3 - 8√3 = 14. Therefore, trapezoid area is (28 + 14)/2 * 2 = (42/2)*2 = 42. Correct. Also check AC perpendicular to CD.

Coordinates:

A(0,0), D(28, 0), B(8√3, 2), C(14 + 8√3, 2).

Vector AC: (14 + 8√3, 2).

Vector CD: D - C = (28 - (14 + 8√3), 0 - 2) = (14 - 8√3, -2).

Dot product: (14 + 8√3)(14 - 8√3) + (2)(-2) = [14² - (8√3)^2] - 4 = [196 - 192] - 4 = 4 - 4 = 0. Correct. Also, angle bisector check.

Vector AB: (8√3, 2). Unit vector: (8√3 / sqrt( (8√3)^2 + 2^2 ), 2 / sqrt( ... )) = (8√3 / sqrt(192 + 4), 2 / sqrt(196)) = (8√3 / 14, 2 / 14) = (4√3 / 7, 1/7).

Unit vector of AD is (1, 0). Sum of unit vectors: (1 + 4√3 / 7, 0 + 1/7) = ( (7 + 4√3)/7, 1/7 ). Direction vector proportional to (7 + 4√3, 1). But AC direction vector is (14 + 8√3, 2) which is 2*(7 + 4√3, 1). Therefore, same direction. Therefore, AC is angle bisector. All conditions satisfied. Therefore, [ACD] = 28. So regardless of the specific values, the area is 28. Therefore, the answer is 28.

But wait, the problem says "compute [triangle ACD]". So 28? But let me check once more. The trapezoid area is 42, and triangle ACD is 28, which is 2/3 of the trapezoid area? Wait, 28 is 2/3 of 42? 42*(2/3) = 28. Yes. But why?

Alternatively, since the trapezoid is divided by AC into triangle ABC and ACD. If [ACD] = 28, then [ABC] = 14. Maybe there's a ratio here. Since AC is an angle bisector, maybe the areas are in some ratio?

Alternatively, since AC bisects angle BAD, and in triangle ABD, but I think we already checked with coordinates. Anyway, since multiple examples confirm that [ACD] = 28, and the derivation seems correct, I think the answer is 28.

But let me check once more with another approach. Let me think about ratios.

Since AC is the angle bisector of angle BAD, and in triangle ABD, but AC meets BD? Wait, no, AC meets BC at point C. Wait, maybe using the angle bisector theorem in triangle ABD? Wait, but point C is not on BD. Maybe not helpful.

Alternatively, since AC is the angle bisector, and AC ⊥ CD, maybe there are some similar triangles?

In triangle ACD, right-angled at C, and angle at A is θ, which is half of angle BAD. Also, in triangle ABC?

Alternatively, since AC bisects angle BAD, and AC ⊥ CD, maybe triangle ACD is related to triangle ABC?

Alternatively, since we found that [ACD] = (h d)/2 and [ABCD] = ( (d + BC)/2 ) * h = 42, and we found that [ACD] = 28, which is 2/3 of the trapezoid area. But why?

Alternatively, since BC = q - p = 7 in the first example, and AD = 14, so BC is half of AD. Therefore, the trapezoid has bases in ratio 2:1. Then area is average of bases times height. If BC is half of AD, then average is (AD + BC)/2 = (14 + 7)/2 = 10.5, times height 4 is 42. Then [ACD] is (h d)/2 = 4*14 / 2 = 28. Which is 2/3 of 42. But why?

Alternatively, since AC divides the trapezoid into two parts, and because of the angle bisector and the perpendicularity, the areas have a specific ratio. But since we have multiple confirmations, I think 28 is correct.

Therefore, I think the answer is \boxed{28}.

**Final Answer**
\boxed{28}

To solve the problem, we are given a trapezoid $ABCD$ with the following properties:

- $AD \parallel BC$
- $\overline{AC} \perp \overline{CD}$
- $\overline{AC}$ bisects $\angle BAD$
- The area of the trapezoid is $[ABCD] = 42$
- We are to find the area of triangle $\triangle ACD$

---

### Step 1: Understand the Geometry

Let’s place the trapezoid in the coordinate plane for clarity:

- Let $A = (0, 0)$
- Since $AD \parallel BC$, and we can assume $AD$ is horizontal, let $D = (d, 0)$
- Let $B = (p, h)$ and $C = (q, h)$, so that $BC$ is also horizontal (same y-coordinate as $B$ and $C$)

This setup ensures that $AD$ and $BC$ are the two bases of the trapezoid, and $AB$ and $CD$ are the legs.

---

### Step 2: Use the Given Conditions

#### (a) $\overline{AC} \perp \overline{CD}$

This implies that the vectors $\vec{AC}$ and $\vec{CD}$ are perpendicular. Using coordinates:

- $\vec{AC} = (q, h)$
- $\vec{CD} = (d - q, -h)$

Their dot product is:

$$
q(d - q) + h(-h) = 0 \Rightarrow qd - q^2 - h^2 = 0 \tag{1}
$$

#### (b) $\overline{AC}$ bisects $\angle BAD$

This implies that the direction vector of $\vec{AC}$ lies between $\vec{AB}$ and $\vec{AD}$. Using vector angle bisector properties, we can derive that:

$$
q = \sqrt{p^2 + h^2} + p \tag{2}
$$

Let’s define $m = \sqrt{p^2 + h^2}$, the length of $\vec{AB}$. Then:

$$
q = m + p \tag{3}
$$

From equation (1), we can also express:

$$
qd = q^2 + h^2 \Rightarrow d = \frac{q^2 + h^2}{q} \tag{4}
$$

Substituting (3) into (4):

$$
d = \frac{(m + p)^2 + h^2}{m + p}
$$

But since $m^2 = p^2 + h^2$, we can simplify this expression further.

---

### Step 3: Use the Area of the Trapezoid

The area of trapezoid $ABCD$ is:

$$
[ABCD] = \frac{1}{2}(AD + BC) \cdot h = \frac{1}{2}(d + (q - p)) \cdot h = 42
$$

From earlier, we found:

$$
d + q - p = 3m \Rightarrow \frac{1}{2}(3m) \cdot h = 42 \Rightarrow \frac{3}{2}mh = 42 \Rightarrow mh = 28 \tag{5}
$$

---

### Step 4: Compute the Area of Triangle $ACD$

Since $\overline{AC} \perp \overline{CD}$, triangle $ACD$ is a right triangle with right angle at $C$. Therefore, its area is:

$$
[\triangle ACD] = \frac{1}{2} \cdot AC \cdot CD
$$

We can express both $AC$ and $CD$ in terms of $m$ and $h$:

- $AC = \sqrt{q^2 + h^2} = \sqrt{(m + p)^2 + h^2} = \sqrt{m^2 + 2mp + p^2 + h^2} = \sqrt{2m^2 + 2mp} = \sqrt{2m(m + p)}$
- From earlier, $CD = \frac{h^2}{q}$, and since $q = m + p$, we get:

$$
CD = \frac{h^2}{m + p}
$$

So the area becomes:

$$
[\triangle ACD] = \frac{1}{2} \cdot \sqrt{2m(m + p)} \cdot \frac{h^2}{m + p} = \frac{1}{2} \cdot \frac{h^2 \sqrt{2m(m + p)}}{m + p}
$$

But this is complicated. Instead, we use a more elegant approach.

From earlier, we found:

$$
[\triangle ACD] = \frac{1}{2} \cdot h \cdot d
$$

And from equation (5), $mh = 28$, and from earlier, $d = 2m$, so:

$$
h \cdot d = h \cdot 2m = 2 \cdot mh = 2 \cdot 28 = 56
$$

Thus:

$$
[\triangle ACD] = \frac{1}{2} \cdot 56 = \boxed{28}
$$

---

### Final Answer

$$
\boxed{28}
$$

Solution:
There are many ways to draw the diagram; one possibility is shown below.  We know that $\angle BAC \cong \angle CAD\cong \angle BCA$ since $\overline{AC}$ bisects $\angle BAD$ and $\overline{AD} || \overline{BC}$.  Therefore $\triangle BAC$ is isosceles.  In the diagram we have added segment $\overline{BE}$ splitting $\triangle BAC$ into two smaller congruent right triangles.  We also know that $\triangle ACD$ is a right triangle by the givens, so we conclude that $\triangle ACD \sim \triangle CEB$ since we already know that $\angle CAD\cong\angle ECB$.  In fact, $\triangle ACD $ is exactly four times the size of $\triangle CEB$ since $AC=2(EC)$.  If we let $[\triangle CEB]=K$, then $[\triangle AEB]=K$ while $[\triangle ACD]=4K$.  Thus $6K=42$, so $K=7$ and $[\triangle ACD]=4K=\boxed{28}$.

[asy]
import olympiad; import graph; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
int randangle = 50;
draw((-5,0)--(5*dir(randangle))--(5,0)--cycle);
path x1 = (5*Cos(randangle),5*Sin(randangle))--(-10,5*Sin(randangle));
path x2 = (-5,0)--(5dir(2*randangle));
pair X = intersectionpoint(x1,x2);
draw((-5,0)--X--(5*dir(randangle)));
draw(rightanglemark((5,0),(5*dir(50)),(-5,0),s=14));
draw(anglemark((5,0),(-5,0),X,18));
draw(anglemark(X,5*dir(50),(-5,0),18));
label("$A$",(-5,0),W); label("$D$",(5,0),E); label("$C$",(5*dir(50)),E);
label("$B$",(X),N);

pair L = foot(X,(-5,0),5*dir(50));
draw(X--L);
draw(rightanglemark(X,L,(-5,0),14)); draw(rightanglemark(X,L,(5*dir(50)),14));
label("$E$",L,SE);
[/asy]