In a right-angled triangle ABC with ∠B = 90°, AB = 8 units and BC = 6 units. Point D is on side AC such that AD:DC = 2:3. Point E is on side BC such that BE:EC = 3:4. If the area of triangle BDE is maximized, what is the ratio of the areas of triangles BDE and ABC?

To solve this problem, let’s break it down step-by-step.

### Step 1: Understanding the triangle dimensions
In the right-angled triangle \(ABC\) with \(\angle B = 90^\circ\), we have:
- \(AB = 8\) units
- \(BC = 6\) units

Using the Pythagorean theorem, we can calculate \(AC\), the hypotenuse:
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units}
\]

### Step 2: Determining the coordinates of points \(A\), \(B\), and \(C\)
To make calculations easier, let’s assign coordinates to points:
- Let \(B\) be at the origin: \(B = (0, 0)\)
- \(A\) will be along the vertical axis, so \(A = (0, 8)\)
- \(C\) will be along the horizontal axis, so \(C = (6, 0)\)

### Step 3: Finding the coordinates of point \(D\)
Point \(D\) is on side \(AC\) such that \(AD:DC = 2:3\). This means \(D\) divides \(AC\) in the ratio \(2:3\).

Using the section formula for a point dividing a line segment in a given ratio \((m:n)\):
\[
D = \left(\frac{m \cdot x_2 + n \cdot x_1}{m + n}, \frac{m \cdot y_2 + n \cdot y_1}{m + n}\right)
\]
Substitute \(A = (0, 8)\), \(C = (6, 0)\), \(m = 2\), and \(n = 3\):
\[
D = \left(\frac{2 \cdot 6 + 3 \cdot 0}{2 + 3}, \frac{2 \cdot 0 + 3 \cdot 8}{2 + 3}\right) = \left(\frac{12}{5}, \frac{24}{5}\right) = \left(2.4, 4.8\right)
\]

### Step 4: Finding the coordinates of point \(E\)
Point \(E\) is on side \(BC\) such that \(BE:EC = 3:4\). This means \(E\) divides \(BC\) in the ratio \(3:4\).

Using the section formula again:
\[
E = \left(\frac{3 \cdot 6 + 4 \cdot 0}{3 + 4}, \frac{3 \cdot 0 + 4 \cdot 0}{3 + 4}\right) = \left(\frac{18}{7}, \frac{0}{7}\right) = \left(2.571, 0\right)
\]

### Step 5: Calculating the area of \(\triangle ABC\)
The area of \(\triangle ABC\) can be calculated using the formula:
\[
\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 6 = 24 \text{ square units}
\]

### Step 6: Calculating the area of \(\triangle BDE\)
To find the area of \(\triangle BDE\), we can use the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates of \(B\), \(D\), and \(E\):
- \(B = (0, 0)\)
- \(D = \left(2.4, 4.8\right)\)
- \(E = \left(2.571, 0\right)\)

\[
\text{Area of } \triangle BDE = \frac{1}{2} \left| 0(4.8 - 0) + 2.4(0 - 0) + 2.571(0 - 4.8) \right|
\]
\[
= \frac{1}{2} \left| 0 + 0 - 2.571 \cdot 4.8 \right| = \frac{1}{2} \times | -12.3088 | = \frac{1}{2} \times 12.3088 = 6.1544 \text{ square units}
\]

### Step 7: Finding the ratio of areas of \(\triangle BDE\) and \(\triangle ABC\)
Now, we can calculate the ratio of the areas:
\[
\text{Ratio} = \frac{\text{Area of } \triangle BDE}{\text{Area of } \triangle ABC} = \frac{6.1544}{24} = 0.2564
\]
In simplest form, this ratio is approximately \( \boxed{\frac{1}{4}} \).

Would you like further details or have any questions?

---

1. How would the solution change if \(AD:DC\) were a different ratio?
2. How can the section formula be applied to find coordinates of dividing points?
3. Why do we use absolute values when finding the area using coordinates?
4. Can the area of triangle \(BDE\) be maximized further with different segment divisions?
5. How would the coordinates approach differ if \(\triangle ABC\) were not right-angled?

**Tip:** In coordinate geometry, always label points and use clear steps for calculations.

Explore the geometry problem involving a right-angled triangle ABC. With side lengths AB = 8 units and BC = 6 units, find the coordinates of points D and E, and calculate the area ratio of triangles BDE and ABC, yielding a maximized area. Suitable for advanced geometry students.

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