Chứng minh PAB = PEI

To prove that \( \angle PAB = \angle PEI \) for the given triangle \( \triangle ABC \) with incircle \( (O) \), we can follow these steps:

1. **Identify Relevant Points**:
- Let \( I \) be the incenter of triangle \( ABC \).
- Points \( D, E, F \) are the points where the altitudes from \( A, B, C \) meet the sides \( BC, CA, AB \), respectively.

2. **Properties of the Incircle**:
- The incircle touches side \( BC \) at a point \( I \). Hence, \( BI \) and \( CI \) are tangent to the incircle. This means that \( AI \) is an angle bisector of \( \angle A \).

3. **Angles Involved**:
- Since \( AI \) is the angle bisector, angles \( \angle PAB \) and \( \angle PEI \) should be properly evaluated.
- By examining \( \triangle AIB \) and \( \triangle AIE \), we can note that if we consider points \( P \) and \( E \) correctly, their angles should have a correspondence given the properties of tangent lines from point \( I \) to the incircle.

4. **Inscribed Angles**:
- By properties of angles inside a circle, we can connect the angles at points \( P \) and \( E \) to show they both subtend the same segment \( AB \) at points \( A \) and \( I \).

5. **Congruence**:
- If we manage to show that triangles \( \triangle PAB \) and \( \triangle PEI \) share equal corresponding angles (due to tangency properties and angle bisector), then we can conclude that \( \angle PAB = \angle PEI \) due to the Equality of Angles in Similar Triangles.

In summary, through properties related to the incircle and the angles they form, we can derive that indeed \( \angle PAB = \angle PEI \). For a complete formal proof, one would generally employ clear notation to demonstrate the congruence of sub-triangles or the inscribed angle theorem as stated.