Let us consider a triangle ABC having BC=5 cm, CA=4cm, AB=3cm, D,E are points on BC such BD = DE= EC, `angleCAE=theta`, then,`AD^(2)` is equal to

Options

A. (a) `52/3`
B. (b) `52/9`
C. (c) `52/7`
D. (d) 52

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the triangle dimensions
We have a triangle ABC with the following dimensions:
- BC = 5 cm
- CA = 4 cm
- AB = 3 cm

### Step 2: Verify if triangle ABC is a right triangle
Using the Pythagorean theorem:
\[ BC^2 = AB^2 + AC^2 \]
Substituting the values:
\[ 5^2 = 3^2 + 4^2 \\ 25 = 9 + 16 \\ 25 = 25 \]
Since the equation holds true, triangle ABC is a right triangle with the right angle at A.

### Step 3: Divide BC into three equal parts
Since BD = DE = EC, we can find the length of each segment:
\[ BD = DE = EC = \frac{BC}{3} = \frac{5}{3} \text{ cm} \]

### Step 4: Apply the cosine rule in triangle ADC
Using the cosine rule:
\[ AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(C) \]
Where:
- \(AC = 4\)
- \(CD = DE + EC = \frac{5}{3} + \frac{5}{3} = \frac{10}{3}\)

### Step 5: Find \(\cos(C)\)
In triangle ABC, we can find \(\cos(C)\):
\[ \cos(C) = \frac{AC}{BC} = \frac{4}{5} \]

### Step 6: Substitute values into the cosine rule
Now substituting the values into the cosine rule equation:
\[ AD^2 = 4^2 + \left(\frac{10}{3}\right)^2 - 2 \cdot 4 \cdot \frac{10}{3} \cdot \frac{4}{5} \]
Calculating each term:
1. \(4^2 = 16\)
2. \(\left(\frac{10}{3}\right)^2 = \frac{100}{9}\)
3. \(2 \cdot 4 \cdot \frac{10}{3} \cdot \frac{4}{5} = \frac{32}{3}\)

### Step 7: Combine the terms
Now we combine all the terms:
\[ AD^2 = 16 + \frac{100}{9} - \frac{32}{3} \]
Convert \(16\) and \(\frac{32}{3}\) to have a common denominator of \(9\):
\[ 16 = \frac{144}{9} \quad \text{and} \quad \frac{32}{3} = \frac{96}{9} \]
Thus:
\[ AD^2 = \frac{144}{9} + \frac{100}{9} - \frac{96}{9} \\ AD^2 = \frac{144 + 100 - 96}{9} \\ AD^2 = \frac{148}{9} \]

### Step 8: Final calculation
Calculating the final value:
\[ AD^2 = \frac{148}{9} = \frac{52}{3} \]

### Conclusion
Thus, the value of \(AD^2\) is:
\[ \boxed{\frac{52}{9}} \]