ABCD is a square. M is the mid-point of AB and N is the mid-point of BC. DM and AN are joined and they meet at O, then which of the following is correct ?

Options

A. OA : OM = 1 : 2
B. AN = MD
C. `angleADM=angleANB`
D. `angleAMD=angleBAN`

Answer

To solve the problem, we need to analyze the square ABCD and the midpoints M and N, then determine the relationship between the segments AN and DM.

### Step-by-Step Solution:

1. **Understanding the Square and Midpoints:**
- Let the vertices of the square ABCD be A(0, 0), B(1, 0), C(1, 1), and D(0, 1).
- The midpoint M of AB is calculated as:
\[ M = \left(\frac{0 + 1}{2}, \frac{0 + 0}{2}\right) = \left(\frac{1}{2}, 0\right) \]
- The midpoint N of BC is calculated as:
\[ N = \left(\frac{1 + 1}{2}, \frac{0 + 1}{2}\right) = \left(1, \frac{1}{2}\right) \]

2. **Finding the Equations of Lines DM and AN:**
- **Line DM:**
- The slope of line DM (from D to M) is:
\[ \text{slope of DM} = \frac{0 - 1}{\frac{1}{2} - 0} = \frac{-1}{\frac{1}{2}} = -2 \]
- The equation of line DM using point D(0, 1):
\[ y - 1 = -2(x - 0) \implies y = -2x + 1 \]

- **Line AN:**
- The slope of line AN (from A to N) is:
\[ \text{slope of AN} = \frac{\frac{1}{2} - 0}{1 - 0} = \frac{1}{2} \]
- The equation of line AN using point A(0, 0):
\[ y - 0 = \frac{1}{2}(x - 0) \implies y = \frac{1}{2}x \]

3. **Finding the Intersection Point O:**
- To find the intersection point O, set the equations of lines DM and AN equal to each other:
\[ -2x + 1 = \frac{1}{2}x \]
- Rearranging gives:
\[ -2x - \frac{1}{2}x = -1 \implies -\frac{5}{2}x = -1 \implies x = \frac{2}{5} \]
- Substitute \(x = \frac{2}{5}\) back into the equation of line AN to find y:
\[ y = \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{5} \]
- Thus, the coordinates of point O are:
\[ O\left(\frac{2}{5}, \frac{1}{5}\right) \]

4. **Finding Lengths AN and DM:**
- **Length of AN:**
- Using the distance formula:
\[ AN = \sqrt{\left(1 - 0\right)^2 + \left(\frac{1}{2} - 0\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \]

- **Length of DM:**
- Using the distance formula:
\[ DM = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(0 - 1\right)^2} = \sqrt{\left(\frac{1}{2}\right)^2 + (-1)^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \]

5. **Conclusion:**
- Since \(AN = DM\), we conclude that:
\[ AN = DM \]
- Therefore, the correct option is that \(AN\) is equal to \(DM\).