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 `AB xx AC// (IA)^(2)` is

Options

A. `((z_(2)-z_(1))(z_(3)-z_(1)))/((z_(4) -z_(1))^2)`
B. `((z_(2) -z_(1))(z_(1)-z_(3)))/((z_(4)-z_(1))^(2))`
C. `((z_(4)-z_(1))^(2))/((z_(2) -z_(1))(z_(3)-z_(1)))`
D. none of these

Explanation

To solve the problem, we need to find the value of \( \frac{AB \cdot AC}{IA^2} \) for the isosceles triangle ABC with given properties. Let's break down the solution step by step.

### Step 1: Understand the Triangle Configuration
Given that triangle ABC is isosceles with \( AC = BC \) and \( \angle CAB = \theta \), we can denote the vertices of the triangle as follows:
- Let \( z_1 \) represent vertex A,
- Let \( z_2 \) represent vertex B,
- Let \( z_3 \) represent vertex C.

### Step 2: Identify the Incenter
The incenter \( z_4 \) of triangle ABC is the point where the angle bisectors of the triangle intersect. Since \( AC = BC \), angles \( \angle IAB \) and \( \angle IAC \) are equal, both being \( \frac{\theta}{2} \).

### Step 3: Use the Angle Bisector Theorem
By the angle bisector theorem, we can express the relationships between the sides and the incenter:
\[ \frac{z_2 - z_1}{z_4 - z_1} = \frac{|z_2 - z_1|}{|z_4 - z_1|} e^{-i \frac{\theta}{2}} \]
\[ \frac{z_3 - z_1}{z_4 - z_1} = \frac{|z_3 - z_1|}{|z_4 - z_1|} e^{-i \frac{\theta}{2}} \]

### Step 4: Set Up the Equations
From the above relationships, we can derive two equations:
1. \( \frac{z_2 - z_1}{z_4 - z_1} = \frac{AB}{IA} e^{-i \frac{\theta}{2}} \)
2. \( \frac{z_3 - z_1}{z_4 - z_1} = \frac{AC}{IA} e^{-i \frac{\theta}{2}} \)

### Step 5: Multiply the Equations
Multiplying these two equations gives:
\[ \frac{(z_2 - z_1)(z_3 - z_1)}{(z_4 - z_1)^2} = \frac{AB \cdot AC}{IA^2} e^{-i \theta} \]

### Step 6: Find the Required Value
Since \( e^{-i \theta} \) is a complex number with modulus 1, we can simplify this to:
\[ \frac{AB \cdot AC}{IA^2} = \frac{|z_2 - z_1| \cdot |z_3 - z_1|}{|z_4 - z_1|^2} \]

### Conclusion
Thus, we find that:
\[ \frac{AB \cdot AC}{IA^2} = \text{some constant value depending on the triangle's dimensions} \]

### Final Answer
The value of \( \frac{AB \cdot AC}{IA^2} \) is a constant determined by the triangle's dimensions and angles.

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