In an isosceles triangle ABC, right angle at B. I is incentre of triangle then, find the ratio area of `DeltaAIB,DeltaBIC and Delta AIC`.

Answer

To find the ratio of the areas of triangles AIB, BIC, and AIC in the isosceles triangle ABC with a right angle at B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Triangle Dimensions**:
- Let the lengths of the equal sides AB and BC be denoted as \( a \).
- Since triangle ABC is a right triangle at B, the length of the hypotenuse AC can be calculated using the Pythagorean theorem:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}. \]

2. **Determine the Inradius (r)**:
- The area \( A \) of triangle ABC can be calculated as:
\[ A = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times a \times a = \frac{a^2}{2}. \]
- The semi-perimeter \( s \) of triangle ABC is:
\[ s = \frac{AB + BC + AC}{2} = \frac{a + a + a\sqrt{2}}{2} = \frac{2a + a\sqrt{2}}{2} = a\left(1 + \frac{\sqrt{2}}{2}\right). \]
- The inradius \( r \) can be calculated using the formula \( r = \frac{A}{s} \):
\[ r = \frac{\frac{a^2}{2}}{a\left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{a}{2\left(1 + \frac{\sqrt{2}}{2}\right)}. \]

3. **Calculate Areas of Triangles AIB, BIC, and AIC**:
- **Area of triangle AIB**:
\[ \text{Area}_{AIB} = \frac{1}{2} \times AB \times r = \frac{1}{2} \times a \times r = \frac{1}{2} \times a \times \frac{a}{2\left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{a^2}{4\left(1 + \frac{\sqrt{2}}{2}\right)}. \]
- **Area of triangle BIC**:
\[ \text{Area}_{BIC} = \frac{1}{2} \times BC \times r = \frac{1}{2} \times a \times r = \frac{a^2}{4\left(1 + \frac{\sqrt{2}}{2}\right)}. \]
- **Area of triangle AIC**:
\[ \text{Area}_{AIC} = \frac{1}{2} \times AC \times r = \frac{1}{2} \times a\sqrt{2} \times r = \frac{1}{2} \times a\sqrt{2} \times \frac{a}{2\left(1 + \frac{\sqrt{2}}{2}\right)} = \frac{a^2\sqrt{2}}{4\left(1 + \frac{\sqrt{2}}{2}\right)}. \]

4. **Find the Ratio of the Areas**:
- Now we can express the ratio of the areas:
\[ \text{Area}_{AIB} : \text{Area}_{BIC} : \text{Area}_{AIC} = \frac{a^2}{4\left(1 + \frac{\sqrt{2}}{2}\right)} : \frac{a^2}{4\left(1 + \frac{\sqrt{2}}{2}\right)} : \frac{a^2\sqrt{2}}{4\left(1 + \frac{\sqrt{2}}{2}\right)}. \]
- Simplifying this gives:
\[ 1 : 1 : \sqrt{2}. \]

### Final Answer:
The ratio of the areas of triangles AIB, BIC, and AIC is \( 1 : 1 : \sqrt{2} \).