I. \(\Delta \text{PAD}\) is similar to \(\Delta \text{PBC}\)
II. \(\angle \text{PAD} + \angle \text{PDA} = \angle \text{PBC} + \angle \text{PCB}\)
Select the answer using the code given below :
II only
To determine which statements about cyclic quadrilateral \(ABCD\) are correct, let's analyze each statement with appropriate reasoning:
- Statement I: \(\Delta \text{PAD}\) is similar to \(\Delta \text{PBC}\)
- For two triangles to be similar, corresponding angles must be equal, or corresponding sides must be in the same ratio.
- In \(\Delta \text{PAD}\) and \(\Delta \text{PBC}\), \( \angle \text{PAD} \) and \( \angle \text{PBC} \) might not necessarily be equal.
- Though they share the common external point \(P\), we do not have any information suggesting that angles or sides are equal or proportional only from cyclic quadrilateral properties.
- Hence, there is no verification from given data that \(\Delta \text{PAD}\) is similar to \(\Delta \text{PBC}\).
- Statement II: \(\angle \text{PAD} + \angle \text{PDA} = \angle \text{PBC} + \angle \text{PCB}\)
- A cyclic quadrilateral's opposite angles are supplementary, that is, their sum is 180°.
- Consider points \(P\), \(A\), and \(D\):
- \(\angle \text{PAD}\) lies along the extension of line \(AB\), and \(\angle \text{PDA}\) is inside the quadrilateral \(ABCD\).
- By the property of opposing cyclic angles, \(\angle \text{PAD} + \angle \text{PDA} = 180^\circ - \angle \text{APD}\) and \(\angle \text{PBC} + \angle \text{PCB} = 180^\circ - \angle \text{CPD}\).
- Since lines \(AB\) and \(CD\) are extended to meet at point \(P\), it ensures that these particular angle pairs add up as stated, thereby making Statement II valid.
Conclusion: After analysis, we find that only Statement II holds true based on geometric properties of cyclic quadrilaterals. Therefore, the correct answer is "II only".
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