A non right angle bisector of a right angled isosceles triangle divide the triangle in those two parts, whose area are in the ratio.

Options

A. `1:1`
B. `1:sqrt(2)`
C. `1:2`
D. `1:sqrt(2)+1`

Answer

To solve the problem of finding the ratio of the areas of two parts into which a non-right angle bisector divides a right-angled isosceles triangle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Triangle**:
- We have a right-angled isosceles triangle, let's denote it as triangle ABC where angle C is the right angle. Thus, angles A and B are both 45 degrees.
- Let the lengths of the equal sides AB and BC be \( x \). Therefore, the hypotenuse AC will be \( \sqrt{2}x \).

2. **Draw the Angle Bisector**:
- We will draw the angle bisector AD from vertex A to side BC. This bisector will divide the triangle into two smaller triangles: ABD and ADC.

3. **Apply the Angle Bisector Theorem**:
- According to the Angle Bisector Theorem, the ratio of the segments created by the bisector on the opposite side is equal to the ratio of the other two sides.
- Thus, we have:
\[ \frac{BD}{DC} = \frac{AB}{AC} = \frac{x}{\sqrt{2}x} = \frac{1}{\sqrt{2}} \]
- Let \( BD = k \) and \( DC = \sqrt{2}k \).

4. **Express the Lengths**:
- From the ratio, we can express \( BD \) and \( DC \) in terms of \( x \):
\[ BD = \frac{x}{1 + \sqrt{2}}, \quad DC = \frac{\sqrt{2}x}{1 + \sqrt{2}} \]

5. **Calculate Areas of Triangles ABD and ADC**:
- The area of triangle ABD can be calculated using the formula:
\[ \text{Area}_{ABD} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times AB \]
Here, the base is \( BD \) and the height is \( AB = x \):
\[ \text{Area}_{ABD} = \frac{1}{2} \times \frac{x}{1 + \sqrt{2}} \times x = \frac{x^2}{2(1 + \sqrt{2})} \]

- Similarly, for triangle ADC:
\[ \text{Area}_{ADC} = \frac{1}{2} \times DC \times AB = \frac{1}{2} \times \frac{\sqrt{2}x}{1 + \sqrt{2}} \times x = \frac{\sqrt{2}x^2}{2(1 + \sqrt{2})} \]

6. **Find the Ratio of Areas**:
- Now, we need the ratio of the areas:
\[ \frac{\text{Area}_{ABD}}{\text{Area}_{ADC}} = \frac{\frac{x^2}{2(1 + \sqrt{2})}}{\frac{\sqrt{2}x^2}{2(1 + \sqrt{2})}} = \frac{1}{\sqrt{2}} \]

7. **Final Result**:
- Therefore, the ratio of the areas of triangles ABD and ADC is:
\[ 1 : \sqrt{2} \]