In a right-angled triangle $ABC$ with $\angle B = 90^\circ$, the length of side $AB$ = $8$ units and the length of side $BC$ = $6$ 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 the problem, we first find the area of triangle $ ABC $ and then the area of triangle $ BDE $ to determine their ratio.

### Step 1: Find the Area of Triangle $ ABC $

Given:
- $ AB = 8 $ units
- $ BC = 6 $ units

The area $ A_{ABC} $ of triangle $ ABC $ can be calculated using the formula for the area of a right triangle:

\[
A_{ABC} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 6 = 24 \text{ square units}
\]

### Step 2: Locate Points $ D $ and $ E $

Next, we find the coordinates of points $ A $, $ B $, and $ C $:
- $ A(0, 6) $
- $ B(0, 0) $
- $ C(8, 0) $

#### Finding Point $ D $

Point $ D $ divides segment $ AC $ in the ratio $ 2:3 $. To find the coordinates of point $ D $, we use the section formula. The coordinates of $ D $ are given by:

\[
D = \left( \frac{3 \cdot 0 + 2 \cdot 8}{2 + 3}, \frac{3 \cdot 6 + 2 \cdot 0}{2 + 3} \right) = \left( \frac{16}{5}, \frac{18}{5} \right)
\]

#### Finding Point $ E $

Point $ E $ divides segment $ BC $ in the ratio $ 3:4 $. The coordinates of point $ E $ are calculated as follows:

\[
E = \left( \frac{4 \cdot 0 + 3 \cdot 8}{3 + 4}, \frac{4 \cdot 0 + 3 \cdot 0}{3 + 4} \right) = \left( \frac{24}{7}, 0 \right)
\]

### Step 3: Area of Triangle $ BDE $

To find the area of triangle $ BDE $, we use the formula for the area of a triangle given by vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:

\[
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]

Substituting the coordinates of points $ B(0, 0) $, $ D\left(\frac{16}{5}, \frac{18}{5}\right) $, and $ E\left(\frac{24}{7}, 0\right) $:

\[
\text{Area}_{BDE} = \frac{1}{2} \left| 0\left(\frac{18}{5}-0\right) + \frac{16}{5}(0-0) + \frac{24}{7}\left(0 - \frac{18}{5}\right) \right|
\]
\[
= \frac{1}{2} \left| \frac{24}{7} \cdot \left(-\frac{18}{5}\right) \right| = \frac{1}{2} \cdot \frac{24 \cdot 18}{7 \cdot 5} = \frac{216}{70} = \frac{108}{35} \text{ square units}
\]

### Step 4: Ratio of the Areas

Now we find the ratio of the areas $ \frac{A_{BDE}}{A_{ABC}} $:

\[
\text{Ratio} = \frac{\text{Area}_{BDE}}{\text{Area}_{ABC}} = \frac{\frac{108}{35}}{24} = \frac{108}{35 \cdot 24} = \frac{108}{840} = \frac{9}{70}
\]

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

\[
\boxed{\frac{9}{70}}
\]

In a right-angled triangle ABC with $\angle B = 90^\circ$, the length of side AB = 8 units and the length of side 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?

Explore the calculation of the area ratio between triangles BDE and ABC in a right triangle where AB = 8 units and BC = 6 units. This solution involves finding points D and E using the section formula and maximizing the area of triangle BDE.

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