Given a triangle ABC with `AB= 2` and `AC=1`. Internal bisector of `angleBAC` intersects BC at D. If AD = BD and `Delta` is the area of triangle ABC, then find the value of `12Delta^2`.

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Understand the Triangle and Given Information
We have a triangle ABC with:
- \( AB = 2 \)
- \( AC = 1 \)
- The internal bisector of angle \( BAC \) intersects \( BC \) at point \( D \).
- It is given that \( AD = BD \).

### Step 2: Analyze the Implications of \( AD = BD \)
Since \( AD = BD \), triangle \( ABD \) is isosceles with \( AB = 2 \) and \( AD = BD \). This implies that angles \( BAD \) and \( ABD \) are equal. Let's denote these angles as \( x \). Therefore, we have:
- \( \angle BAD = x \)
- \( \angle ABD = x \)
- \( \angle DAC = \angle BAC - \angle BAD = 2x \)

### Step 3: Find the Third Angle
The sum of angles in triangle \( ABC \) is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
Substituting the angles we have:
\[ 2x + x + \angle C = 180^\circ \]
This simplifies to:
\[ \angle C = 180^\circ - 3x \]

### Step 4: Apply the Sine Rule
Using the sine rule in triangle \( ABC \):
\[ \frac{AB}{\sin C} = \frac{AC}{\sin B} \]
Substituting the known values:
\[ \frac{2}{\sin(180^\circ - 3x)} = \frac{1}{\sin x} \]
Since \( \sin(180^\circ - \theta) = \sin \theta \):
\[ \frac{2}{\sin 3x} = \frac{1}{\sin x} \]
Cross-multiplying gives:
\[ 2 \sin x = \sin 3x \]

### Step 5: Use the Sine Triple Angle Identity
Using the identity \( \sin 3x = 3 \sin x - 4 \sin^3 x \):
\[ 2 \sin x = 3 \sin x - 4 \sin^3 x \]
Rearranging gives:
\[ 4 \sin^3 x - \sin x = 0 \]
Factoring out \( \sin x \):
\[ \sin x (4 \sin^2 x - 1) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \) (not possible for a triangle)
2. \( 4 \sin^2 x - 1 = 0 \)

### Step 6: Solve for \( \sin x \)
From \( 4 \sin^2 x - 1 = 0 \):
\[ 4 \sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \implies \sin x = \frac{1}{2} \]

### Step 7: Calculate the Area of Triangle \( ABC \)
The area \( \Delta \) of triangle \( ABC \) can be calculated using:
\[ \Delta = \frac{1}{2} \cdot AB \cdot AC \cdot \sin A \]
Where \( A = 2x \):
\[ \Delta = \frac{1}{2} \cdot 2 \cdot 1 \cdot \sin(2x) \]
Using the identity \( \sin(2x) = 2 \sin x \cos x \):
\[ \Delta = \frac{1}{2} \cdot 2 \cdot 1 \cdot 2 \cdot \sin x \cdot \cos x \]
Substituting \( \sin x = \frac{1}{2} \):
\[ \Delta = 2 \cdot \frac{1}{2} \cdot \cos x \]
Calculating \( \cos x \):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \frac{1}{4} = \frac{3}{4} \implies \cos x = \frac{\sqrt{3}}{2} \]
Thus,
\[ \Delta = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Step 8: Find \( 12 \Delta^2 \)
Now we calculate \( 12 \Delta^2 \):
\[ \Delta^2 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Thus,
\[ 12 \Delta^2 = 12 \cdot \frac{3}{4} = 9 \]

### Final Answer
The value of \( 12 \Delta^2 \) is \( \boxed{9} \).