In an Agrad plane `z_(1),z_(2) and z_(3)` are, respectively, the vertices of an isosceles trinagle ABC with AC= BC and `/_CAB = theta`. If `z_(4)` is incentre of triangle, then
The value of `(z_(2)-z_(1))^(2)tan theta tan theta //2` is

Options

A. `(z_(1) + z_(2)-2z_(3))`
B. `(z_(1)+z_(2)-z_(3))(z_(1) +z_(2)-z_(4))`
C. `-(z_(1) +z_(2)-2z_(3))(z_(1)+z_(2)-2z_(4))`
D. `z_(4)=sqrt(z_(2)z_(3))`

Explanation

To solve the problem, we need to find the value of \((z_2 - z_1)^2 \tan \theta \tan \frac{\theta}{2}\) given the vertices of an isosceles triangle \(ABC\) with \(AC = BC\) and \(\angle CAB = \theta\). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Geometry
We have an isosceles triangle \(ABC\) where \(AC = BC\) and \(\angle CAB = \theta\). Let:
- \(z_1\) represent the vertex \(A\),
- \(z_2\) represent the vertex \(B\),
- \(z_3\) represent the vertex \(C\).

### Step 2: Define the Coordinates
Assume the coordinates of the vertices in the complex plane:
- \(z_1 = 0\) (for simplicity, we place \(A\) at the origin),
- \(z_2 = b\) (let \(B\) be at \(b\) on the real axis),
- \(z_3 = c\) (the coordinates of \(C\) will depend on the angle \(\theta\)).

Since \(AC = BC\), we can express \(C\) in terms of \(b\) and \(\theta\):
- The coordinates of \(C\) can be represented as \(z_3 = \frac{b}{2} + i \cdot h\), where \(h\) is the height from \(C\) to the base \(AB\).

### Step 3: Calculate the Height \(h\)
Using trigonometric relationships:
- The height \(h\) can be expressed as:
\[ h = AC \sin \theta = \frac{b}{2} \tan \frac{\theta}{2} \]

### Step 4: Find \(z_2 - z_1\)
Now, calculate \(z_2 - z_1\):
\[ z_2 - z_1 = b - 0 = b \]

### Step 5: Calculate \((z_2 - z_1)^2\)
Now, square the difference:
\[ (z_2 - z_1)^2 = b^2 \]

### Step 6: Find \(\tan \theta\) and \(\tan \frac{\theta}{2}\)
Using the definitions of tangent:
\[ \tan \theta = \frac{h}{\frac{b}{2}} = \frac{2h}{b} \]
\[ \tan \frac{\theta}{2} = \frac{h}{\frac{b}{2}} = \frac{2h}{b} \]

### Step 7: Substitute into the Expression
Now substitute these values into the expression:
\[ (z_2 - z_1)^2 \tan \theta \tan \frac{\theta}{2} = b^2 \left(\frac{2h}{b}\right) \left(\frac{2h}{b}\right) \]
\[ = b^2 \cdot \frac{4h^2}{b^2} = 4h^2 \]

### Step 8: Final Expression
Thus, the value of \((z_2 - z_1)^2 \tan \theta \tan \frac{\theta}{2}\) simplifies to:
\[ 4h^2 \]

### Conclusion
The final value of \((z_2 - z_1)^2 \tan \theta \tan \frac{\theta}{2}\) is \(4h^2\).