If AD is median of triangle ABC and P is a point on AC such that area `(DeltaADP) : area (Delta ABD) = 2 :3.`, then area `(DeltaPDC) : area (DeltaABC)` is equal to:

Options

A. `1 : 5`
B. `5 : 1`
C. `1 :6`
D. `6 : 1`

Answer

To solve the problem, we need to find the ratio of the area of triangle PDC to the area of triangle ABC given that AD is a median of triangle ABC and the ratio of the areas of triangles ADP and ABD is 2:3.

### Step-by-Step Solution:

1. **Understanding the Triangle and Median**:
- Let triangle ABC have a median AD. This means that D is the midpoint of side BC.
- Therefore, the area of triangle ABD is equal to the area of triangle ADC.

2. **Setting Up the Areas**:
- Let the area of triangle ABD be denoted as \( S \).
- Since AD is a median, the area of triangle ADC is also \( S \).
- The total area of triangle ABC is \( 2S \) (since \( Area(ABC) = Area(ABD) + Area(ADC) \)).

3. **Using the Given Ratio**:
- We know from the problem statement that:
\[ \frac{Area(ADP)}{Area(ABD)} = \frac{2}{3} \]
- Let the area of triangle ADP be \( A_{ADP} \). Then:
\[ A_{ADP} = \frac{2}{3} \cdot S \]

4. **Finding the Area of Triangle PDC**:
- The area of triangle ABD can be expressed as:
\[ Area(ABD) = Area(ADP) + Area(PDC) \]
- Therefore, we can write:
\[ S = A_{ADP} + Area(PDC) \]
- Substituting \( A_{ADP} \):
\[ S = \frac{2}{3}S + Area(PDC) \]
- Rearranging gives:
\[ Area(PDC) = S - \frac{2}{3}S = \frac{1}{3}S \]

5. **Finding the Ratio of Areas**:
- Now we need to find the ratio of the area of triangle PDC to the area of triangle ABC:
\[ \frac{Area(PDC)}{Area(ABC)} = \frac{\frac{1}{3}S}{2S} = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \]

### Conclusion:
The ratio of the area of triangle PDC to the area of triangle ABC is:
\[ \frac{Area(PDC)}{Area(ABC)} = \frac{1}{6} \]