Quadrilateral $ABCD$ satisfies $AB = 8, BC = 5, CD = 17, DA = 10$. Let $E$ be the intersection of $AC$ and $BD$. Suppose $BE : ED = 1 : 2$. Find the area of $ABCD$.
1. **Given Data and Setup:**
   - Quadrilateral \(ABCD\) with sides \(AB = 8\), \(BC = 5\), \(CD = 17\), and \(DA = 10\).
   - \(E\) is the intersection of diagonals \(AC\) and \(BD\).
   - Ratio \(BE : ED = 1 : 2\).

2. **Area Ratios and Sine Rule:**
   - Since \(BE : ED = 1 : 2\), the areas of triangles \(\triangle ABE\) and \(\triangle CDE\) are in the ratio \(1 : 2\).
   - Therefore, \([ABE] : [CDE] = 1 : 2\).
   - Using the sine rule in triangles \(\triangle ABE\) and \(\triangle CDE\), we have:
     \[
     \frac{[ABE]}{[CDE]} = \frac{1}{2} = \frac{AB \cdot BE \cdot \sin \angle ABE}{CD \cdot ED \cdot \sin \angle CDE}
     \]
     Given \(BE : ED = 1 : 2\), we can simplify this to:
     \[
     \frac{8 \cdot BE \cdot \sin \angle ABE}{17 \cdot 2BE \cdot \sin \angle CDE} = \frac{1}{2}
     \]
     Simplifying, we get:
     \[
     8 \sin \angle ABE = 17 \sin \angle CDE
     \]
     Let \(\sin \angle ABE = \sin B\) and \(\sin \angle CDE = \sin D\), then:
     \[
     8 \sin B = 17 \sin D \quad \text{(1)}
     \]

3. **Law of Cosines:**
   - Using the law of cosines in \(\triangle ABC\) and \(\triangle ADC\):
     \[
     AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B
     \]
     \[
     AC^2 = 8^2 + 5^2 - 2 \cdot 8 \cdot 5 \cdot \cos B = 64 + 25 - 80 \cos B = 89 - 80 \cos B
     \]
     Similarly, for \(\triangle ADC\):
     \[
     AC^2 = AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos D
     \]
     \[
     AC^2 = 10^2 + 17^2 - 2 \cdot 10 \cdot 17 \cdot \cos D = 100 + 289 - 340 \cos D = 389 - 340 \cos D
     \]
     Equating the two expressions for \(AC^2\):
     \[
     89 - 80 \cos B = 389 - 340 \cos D \quad \text{(2)}
     \]

4. **Solving the Equations:**
   - From equation (1), we have:
     \[
     \sin D = \frac{8}{17} \sin B
     \]
   - Substitute \(\sin D\) into equation (2):
     \[
     89 - 80 \cos B = 389 - 340 \cos D
     \]
     Using \(\cos^2 D = 1 - \sin^2 D\):
     \[
     \cos D = \sqrt{1 - \left(\frac{8}{17} \sin B\right)^2}
     \]
     Substitute back:
     \[
     89 - 80 \cos B = 389 - 340 \sqrt{1 - \left(\frac{8}{17} \sin B\right)^2}
     \]
     Simplify and solve for \(\cos B\):
     \[
     15 + 4 \cos B = \sqrt{17^2 - (8 \sin B)^2}
     \]
     \[
     15 + 4 \cos B = \sqrt{289 - 64 \sin^2 B}
     \]
     \[
     15 + 4 \cos B = \sqrt{289 - 64 \left(1 - \cos^2 B\right)}
     \]
     \[
     15 + 4 \cos B = \sqrt{225 + 64 \cos^2 B}
     \]
     Square both sides:
     \[
     (15 + 4 \cos B)^2 = 225 + 64 \cos^2 B
     \]
     \[
     225 + 120 \cos B + 16 \cos^2 B = 225 + 64 \cos^2 B
     \]
     \[
     120 \cos B + 16 \cos^2 B = 64 \cos^2 B
     \]
     \[
     120 \cos B = 48 \cos^2 B
     \]
     \[
     48 \cos^2 B - 120 \cos B = 0
     \]
     \[
     \cos B (48 \cos B - 120) = 0
     \]
     \[
     \cos B = 0 \quad \text{or} \quad \cos B = \frac{120}{48} = 2.5 \quad \text{(not possible)}
     \]
     Therefore, \(\cos B = 0\), which implies \(\angle B = 90^\circ\).

5. **Area Calculation:**
   - Since \(\angle B = 90^\circ\), \(\triangle ABC\) is a right triangle.
   - Area of \(\triangle ABC\):
     \[
     [ABC] = \frac{1}{2} \cdot AB \cdot BC = \frac{1}{2} \cdot 8 \cdot 5 = 20
     \]
   - Given the ratio of areas, \([ABD] = 3 \times [ABC]\):
     \[
     [ABCD] = 3 \times 20 = 60
     \]

The final answer is \(\boxed{60}\).