In the `DeltaABC`, points M and N respectively lie on side AB and AC such that area of triangle ABC is double than area of trapezium BMNC, The ratio `AM:MB` is.

Options

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

Answer

To solve the problem, we need to find the ratio \( AM:MB \) given that the area of triangle \( ABC \) is double that of trapezium \( BMNC \).

### Step-by-Step Solution:

1. **Understanding the Areas**:
- Let the area of triangle \( ABC \) be \( A \).
- The area of trapezium \( BMNC \) is given to be half of the area of triangle \( ABC \). Therefore, if \( A \) is the area of triangle \( ABC \), then the area of trapezium \( BMNC \) is \( \frac{A}{2} \).

2. **Area Relationships**:
- The area of triangle \( ABC \) can be expressed as the sum of the areas of triangle \( ABM \) and triangle \( ACN \) plus the area of trapezium \( BMNC \).
- Thus, we can write:
\[ A = \text{Area of } ABM + \text{Area of } ACN + \text{Area of } BMNC \]
- Since we know that the area of trapezium \( BMNC \) is \( \frac{A}{2} \), we can substitute this into the equation:
\[ A = \text{Area of } ABM + \text{Area of } ACN + \frac{A}{2} \]

3. **Simplifying the Area Equation**:
- Rearranging gives:
\[ A - \frac{A}{2} = \text{Area of } ABM + \text{Area of } ACN \]
- This simplifies to:
\[ \frac{A}{2} = \text{Area of } ABM + \text{Area of } ACN \]

4. **Using Ratios**:
- Let \( AM = x \) and \( MB = y \). Therefore, \( AB = x + y \).
- The area of triangle \( ABM \) can be expressed in terms of \( x \) and \( y \):
\[ \text{Area of } ABM = \frac{1}{2} \times AB \times h_{ABM} = \frac{1}{2} \times (x + y) \times h_{ABM} \]
- The area of triangle \( ACN \) can be expressed similarly.

5. **Finding the Ratio**:
- From the area relationships we have established, we can derive that:
\[ \frac{\text{Area of } ABM}{\text{Area of } ABC} = \frac{x}{x+y} \]
- Since we know that the total area of \( ABM \) and \( ACN \) is \( \frac{A}{2} \), we can set up the ratio:
\[ \frac{x}{x+y} + \frac{y}{x+y} = 1 \]
- This leads us to find that \( \frac{AM}{MB} = \frac{1}{1} \) or \( 1:1 \).

6. **Final Calculation**:
- To find the specific ratio \( AM:MB \), we can use the relationship derived from the areas:
\[ AM:MB = 1:\sqrt{2}-1 \]
- Rationalizing gives us:
\[ AM:MB = \sqrt{2} + 1 : 1 \]

### Conclusion:
The ratio \( AM:MB \) is \( \sqrt{2} + 1 : 1 \).

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