In a triangle ABC, a point D is chosen on BC such that BD : DC = 2 : 5. Let P be a point on the circumcircleABC such that `anglePDB = angleBAC`. Then PD : PC is :-

Options

A. `sqrt2,sqrt5`
B. `2:5`
C. `2:7`
D. `sqrt2:sqrt7`

Answer

To solve the problem, we need to find the ratio \( PD : PC \) in triangle \( ABC \) with a point \( D \) on \( BC \) such that \( BD : DC = 2 : 5 \) and a point \( P \) on the circumcircle of triangle \( ABC \) such that \( \angle PDB = \angle BAC \).

### Step-by-Step Solution:

1. **Understanding the Given Ratios**:
We are given that \( BD : DC = 2 : 5 \). Let us denote \( BD = 2k \) and \( DC = 5k \) for some positive value \( k \). Therefore, the total length of \( BC \) is:
\[ BC = BD + DC = 2k + 5k = 7k. \]

2. **Identifying Angles**:
We know that \( \angle PDB = \angle BAC \). Let’s denote \( \angle BAC = \theta \). Therefore, \( \angle PDB = \theta \).

3. **Finding Other Angles**:
Let \( \angle PCD = \alpha \). Since \( \angle PDC + \angle PDB = 180^\circ \) (linear pair), we have:
\[ \angle PDC = 180^\circ - \theta. \]

4. **Using the Angle Sum Property**:
In triangle \( PDC \), the sum of angles is \( 180^\circ \):
\[ \angle DPC + \angle PCD + \angle PDC = 180^\circ. \]
Substituting the known angles:
\[ \angle DPC + \alpha + (180^\circ - \theta) = 180^\circ. \]
Simplifying gives:
\[ \angle DPC = \theta - \alpha. \]

5. **Using the Property of Angles in the Same Segment**:
Since \( \angle BAC = \angle BPC \) (angles in the same segment), we have:
\[ \angle BPC = \theta. \]

6. **Finding \( \angle BPD \)**:
The angle \( \angle BPD \) can be expressed as:
\[ \angle BPD = \theta - (180^\circ - \theta) = 2\theta - 180^\circ. \]

7. **Establishing Similarity of Triangles**:
Triangles \( PBC \) and \( PDC \) are similar because they share angle \( P \) and have equal angles \( \angle BPC \) and \( \angle PDC \). Therefore, we can write:
\[ \frac{PC}{PD} = \frac{BC}{BD}. \]

8. **Substituting Known Values**:
We know \( BC = 7k \) and \( BD = 2k \). Thus:
\[ \frac{PC}{PD} = \frac{7k}{2k} = \frac{7}{2}. \]

9. **Finding the Ratio \( PD : PC \)**:
Taking the reciprocal gives:
\[ PD : PC = \frac{2}{7}. \]

### Final Answer:
Thus, the ratio \( PD : PC \) is:
\[ \boxed{2 : 7}. \]