We are tasked with finding the ratio of the areas of triangles $\triangle BDE$ and $\triangle ABC$, where certain conditions on the positions of points $D$ and $E$ apply. Let’s go through this step by step, applying geometric concepts and logical reasoning.

1. Step 1: Analyze Triangle ABC

• $\triangle ABC$ is a right-angled triangle with $\angle B = 90^\circ$.
• The lengths of sides are:
  • $AB = 8$ units
  • $BC = 6$ units
  • We can calculate the hypotenuse $AC$ using the Pythagorean theorem:

$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units}.$

2. Step 2: Set up the partition on sides AC and BC

• Point $D$ divides side $AC$ in the ratio $AD:DC = 2:3$, meaning $AD = \frac{2}{5}AC$ and $DC = \frac{3}{5}AC$.

  • $AD = \frac{2}{5} \times 10 = 4$ units.
  • $DC = \frac{3}{5} \times 10 = 6$ units.

• Point $E$ divides side $BC$ in the ratio $BE:EC = 3:4$, meaning $BE = \frac{3}{7}BC$ and $EC = \frac{4}{7}BC$.

  • $BE = \frac{3}{7} \times 6 = \frac{18}{7} \approx 2.57$ units.
  • $EC = \frac{4}{7} \times 6 = \frac{24}{7} \approx 3.43$ units.

3. Step 3: Area of triangle ABC

The area of $\triangle ABC$ is given by:

$\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 6 = 24 \text{ square units}.$

4. Step 4: Maximize the area of triangle BDE

To maximize the area of triangle $\triangle BDE$, we use the geometric fact that the area of any triangle is half the product of its base and height. Here, $BE$ is part of side $BC$, and point $D$ divides side $AC$. The area will depend on how these points divide the triangle, but it's maximized when both $D$ and $E$ lie in such a way that the triangle is stretched optimally within $\triangle ABC$.

5. Step 5: Calculate the ratio of areas

The area of $\triangle BDE$ is a fraction of the area of $\triangle ABC$, and this fraction depends on the positions of $D$ and $E$. Since $D$ divides $AC$ in the ratio 2:3 and $E$ divides $BC$ in the ratio 3:4, the area of $\triangle BDE$ can be approximated by multiplying these ratios:

$\text{Ratio of areas} = \left( \frac{2}{5} \right) \times \left( \frac{3}{7} \right) = \frac{6}{35}.$

Thus, the ratio of the areas of triangles $\triangle BDE$ and $\triangle ABC$ is:

$\boxed{\frac{6}{35}}.$

Final Result:

The ratio of the areas of triangles $\triangle BDE$ and $\triangle ABC$ is $\frac{6}{35}$.

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Would you like to explore any steps in more detail or clarify any concepts?

Here are 5 related questions you might want to consider:

1. How do we find the centroid of a triangle, and how could it help in area maximization?
2. What is the significance of the ratio $AD:DC = 2:3$ in determining the area?
3. How would the problem change if point $E$ divided side $BC$ in a different ratio?
4. Can the area of triangle BDE be further maximized by changing other parameters?
5. How can we apply similar triangle properties to calculate areas more directly?

Tip: When dealing with ratios in geometry, consider using area proportions based on side divisions, as they can simplify complex area comparisons.