In a triangle OAB, E is the midpoint of OB and D is a point on AB such that `AD:DB=2:1`. IF OD and AE intersects at P, then ratio of `(OP)/(PD)` is equal to

Options

A. `3:2`
B. `2:3`
C. `3:4`
D. `4:3`

Answer

To solve the problem, we need to find the ratio of \( OP \) to \( PD \) in triangle \( OAB \) where \( E \) is the midpoint of \( OB \) and \( D \) divides \( AB \) in the ratio \( 2:1 \).

### Step-by-Step Solution:

1. **Define the Points Using Vectors**:
Let the position vectors of points \( O \), \( A \), and \( B \) be represented as \( \vec{O} = \vec{0} \), \( \vec{A} = \vec{a} \), and \( \vec{B} = \vec{b} \).

2. **Find the Position Vector of Point \( E \)**:
Since \( E \) is the midpoint of \( OB \):
\[ \vec{E} = \frac{\vec{O} + \vec{B}}{2} = \frac{\vec{0} + \vec{b}}{2} = \frac{\vec{b}}{2} \]

3. **Find the Position Vector of Point \( D \)**:
Point \( D \) divides \( AB \) in the ratio \( 2:1 \). Using the section formula:
\[ \vec{D} = \frac{2\vec{B} + 1\vec{A}}{2 + 1} = \frac{2\vec{b} + \vec{a}}{3} \]

4. **Express the Points \( P \) on Lines \( AE \) and \( OD \)**:
Let \( P \) divide \( AE \) in the ratio \( \lambda : 1 \). Then:
\[ \vec{P} = \frac{\lambda \vec{E} + 1 \vec{A}}{\lambda + 1} = \frac{\lambda \left(\frac{\vec{b}}{2}\right) + \vec{a}}{\lambda + 1} = \frac{\frac{\lambda \vec{b}}{2} + \vec{a}}{\lambda + 1} \]

5. **Express \( P \) on Line \( OD \)**:
Let \( P \) divide \( OD \) in the ratio \( \gamma : 1 \):
\[ \vec{P} = \frac{\gamma \vec{D} + 1 \vec{O}}{\gamma + 1} = \frac{\gamma \left(\frac{2\vec{b} + \vec{a}}{3}\right) + \vec{0}}{\gamma + 1} = \frac{\frac{2\gamma \vec{b} + \gamma \vec{a}}{3}}{\gamma + 1} \]

6. **Equate the Two Expressions for \( \vec{P} \)**:
Set the two expressions for \( \vec{P} \) equal to each other:
\[ \frac{\frac{\lambda \vec{b}}{2} + \vec{a}}{\lambda + 1} = \frac{\frac{2\gamma \vec{b} + \gamma \vec{a}}{3}}{\gamma + 1} \]

7. **Cross Multiply to Eliminate the Denominators**:
Cross-multiplying gives:
\[ \left(\frac{\lambda \vec{b}}{2} + \vec{a}\right)(\gamma + 1) = \left(2\gamma \vec{b} + \gamma \vec{a}\right)(\lambda + 1) \]

8. **Expand and Rearrange**:
Expanding both sides and rearranging will lead to:
\[ \frac{\lambda \gamma \vec{b}}{2} + \lambda \vec{a} + \gamma \frac{\vec{b}}{2} + \vec{a} = 2\lambda \gamma \vec{b} + \lambda \gamma \vec{a} + 2\gamma \vec{b} + \gamma \vec{a} \]

9. **Solve for \( \lambda \) and \( \gamma \)**:
After simplifying, we will find values for \( \lambda \) and \( \gamma \).

10. **Find the Ratio \( \frac{OP}{PD} \)**:
From the expressions for \( OP \) and \( PD \):
\[ OP = \frac{3}{5} OD \quad \text{and} \quad PD = \frac{2}{5} OD \]
Thus, the ratio:
\[ \frac{OP}{PD} = \frac{\frac{3}{5} OD}{\frac{2}{5} OD} = \frac{3}{2} \]

### Final Answer:
The ratio \( \frac{OP}{PD} \) is \( \frac{3}{2} \).