If AD is a median of a `Delta ABC` and P is a point on AC such that ar`(Delta ADP)` : ar `(Delta ABD)` = 2:3, then ar `(Delta PDC)` : ar `(Delta ABC)` is .

Options

A. `1:6`
B. `1:5`
C. `2:5`
D. `3:5`

Answer

To solve the problem, we will follow these steps:

1. **Understand the Given Information**: We have triangle ABC with AD as a median. This means D is the midpoint of BC. We also know that the area of triangle ADP is in the ratio of 2:3 with the area of triangle ABD.

2. **Assign Variables**: Let's denote the area of triangle ABD as \( X \). Since D is the midpoint, the area of triangle ACD will also be \( X \).

3. **Calculate the Area of Triangle ADP**: Given that the area of triangle ADP to the area of triangle ABD is \( 2:3 \), we can express the area of triangle ADP as:
\[ \text{Area of } \triangle ADP = \frac{2}{3} \times X = \frac{2X}{3} \]

4. **Calculate the Area of Triangle ADC**: The area of triangle ADC is equal to the area of triangle ABD, which we have already defined as \( X \).

5. **Find the Area of Triangle PDC**: To find the area of triangle PDC, we can use the relationship:
\[ \text{Area of } \triangle ADC = \text{Area of } \triangle ADP + \text{Area of } \triangle PDC \]
Therefore, we can express the area of triangle PDC as:
\[ \text{Area of } \triangle PDC = \text{Area of } \triangle ADC - \text{Area of } \triangle ADP \]
Substituting the values we have:
\[ \text{Area of } \triangle PDC = X - \frac{2X}{3} = \frac{3X}{3} - \frac{2X}{3} = \frac{X}{3} \]

6. **Calculate the Area of Triangle ABC**: The area of triangle ABC is the sum of the areas of triangles ABD and ACD:
\[ \text{Area of } \triangle ABC = \text{Area of } \triangle ABD + \text{Area of } \triangle ACD = X + X = 2X \]

7. **Find the Ratio of Areas**: Now we can find the ratio of the area of triangle PDC to the area of triangle ABC:
\[ \text{Ratio} = \frac{\text{Area of } \triangle PDC}{\text{Area of } \triangle ABC} = \frac{\frac{X}{3}}{2X} = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \]

Thus, the final answer is:
\[ \text{ar}(\Delta PDC) : \text{ar}(\Delta ABC) = 1 : 6 \]