AB and CD are two chords of a circle meeting externally at P. Then, which of the following is/are correct?
I.PAxPD = PCxPB. traingle PAC and traingle PDB are similar.
Select the correct answer using the codes given below.

Options

A. Only I
B. Only II
C. Both I and II
D. Neither I nor II

Answer

To solve the problem, we need to analyze the statements given about the two chords AB and CD that meet externally at point P.

### Step-by-Step Solution:

1. **Understanding the Chords and Point of Intersection**:
- We have two chords, AB and CD, intersecting externally at point P.
- The segments created by the intersection are PA, PB, PC, and PD.

2. **Analyzing the First Statement**:
- The first statement claims that \( PA \times PD = PC \times PB \).
- According to the theorem of intersecting chords, when two chords intersect, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
- Therefore, we can express this as:
\[ PA \times PB = PC \times PD \]
- Rearranging gives us:
\[ PA \times PD = PC \times PB \]
- This confirms that the first statement is **correct**.

3. **Analyzing the Second Statement**:
- The second statement claims that triangles PAC and PDB are similar.
- For two triangles to be similar, they must have equal corresponding angles.
- Here, we note that:
- Angle PAC is not necessarily equal to angle PDB unless the lines AC and BD are parallel.
- Since we do not have any information suggesting that AC is parallel to BD, we cannot conclude that the triangles are similar.
- Thus, the second statement is **incorrect**.

4. **Conclusion**:
- The first statement is correct: \( PA \times PD = PC \times PB \).
- The second statement is incorrect: triangles PAC and PDB are not necessarily similar.
- Therefore, the correct answer is that only the first statement is true.

### Final Answer:
- The correct answer is: **Only the first statement is correct.**