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We are given triangle \( ABC \) with side lengths \( AB = 18 \), \( BC = 24 \), and \( CA = 20 \). Additionally, points \( D \) and \( E \) are placed on segments \( AB \) and \( BC \), respectively, such that \( AD = 15 \) and \( EC = 20 \). The lines \( AE \) and \( DC \) intersect at point \( F \), and the problem asks to compute:

\[
\left(\text{area of } \triangle DEF\right)^2 \cdot \frac{7^3 \cdot 6^3}{5^3}
\]

We will first compute the area of triangle \( DEF \) using geometry or a proportionality theorem (like Ceva's or Menelaus' theorem) and then apply the expression.

### Step 1: Use Menelaus' Theorem
Menelaus' theorem can help us find relationships between the segments formed by the intersections of the triangle's sides and the transversals \( AE \) and \( DC \). Using this, we can calculate ratios and then the area of triangle \( DEF \).

### Step 2: Calculate the Area of Triangle ABC
We can first find the area of triangle \( ABC \) using Heron's formula:

The semi-perimeter \( s \) of \( \triangle ABC \) is:

\[
s = \frac{AB + BC + CA}{2} = \frac{18 + 24 + 20}{2} = 31
\]

The area \( A \) of \( \triangle ABC \) is:

\[
A = \sqrt{s(s - AB)(s - BC)(s - CA)} = \sqrt{31(31 - 18)(31 - 24)(31 - 20)}
\]

\[
A = \sqrt{31 \times 13 \times 7 \times 11}
\]

Let me calculate this value for you.The area of triangle \( ABC \) is approximately \( 176.16 \, \text{square units} \).

### Step 3: Proportions and Area of Triangle DEF
Since \( D \) and \( E \) divide sides \( AB \) and \( BC \) in the ratios \( \frac{AD}{DB} = \frac{15}{3} = 5 \) and \( \frac{EC}{EB} = \frac{20}{4} = 5 \), triangles \( ABC \) and \( DEF \) are similar by Ceva's theorem. The ratio of their areas will be related to the square of the side ratios.

The area of triangle \( DEF \) is scaled by a factor of \( \frac{1}{5^2} = \frac{1}{25} \) from triangle \( ABC \).

\[
\text{Area of } DEF = \frac{1}{25} \times 176.16 \approx 7.05
\]

### Step 4: Apply the Formula
We now compute:

\[
\left(\text{area of } \triangle DEF\right)^2 \cdot \frac{7^3 \cdot 6^3}{5^3}
\]

First, let's calculate the value for this expression.The computed value of the expression \(\left(\text{area of } \triangle DEF\right)^2 \cdot \frac{7^3 \cdot 6^3}{5^3}\) is approximately \( 29,427.52 \).

Do you want further details on the process or any clarification?

Here are some related questions:
1. What is Menelaus' theorem, and how is it used in triangle geometry?
2. Can the area of a triangle be found using only its side lengths?
3. How do the areas of two similar triangles relate to the ratios of their sides?
4. How can Ceva's theorem be applied to solve geometric problems?
5. What are other methods to find the area of a triangle besides Heron's formula?

**Tip:** In geometry problems, identifying and using proportionality theorems like Menelaus' or Ceva's can simplify complex intersections and area calculations.

Let ABC be a triangle with AB = 18, BC = 24, and CA = 20. D is placed on AB such that AD = 15. E is placed on BC such that EC = 20. Call the intersection of lines AE and DC point F. Compute [area of DEF]^2 · 7^3 · 6^3 / 5^3

This problem explores geometry concepts with triangle ABC, where points D and E divide the sides. By using Menelaus' and Ceva's theorems, and applying Heron's formula, we compute the area of triangle DEF and evaluate the given expression. Suitable for Grades 10-12.

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