To solve the problem, we need to approach it step by step, following the instructions given.

### a. Prove that \( BD \) is the angle bisector of \( \angle ABC \).

Given \( \triangle ABC \) is a right triangle at \( A \) with \( N \) on \( BC \) such that \( BN = BA \) and \( D \) lies on line \( AN \) with \( BD \perp AN \).

1. Use the property of angles in right triangles. Since \( A \) is the right angle, \( \angle ABC = 90^\circ - \angle ACB \).
2. Since \( BN = BA \), \( \triangle ABN \) is isosceles with \( \angle BAN = \angle BNA \).
3. By angle properties and the fact that \( D \) lies on \( AN \), show that \( BD \) divides \( \angle ABC \) into two equal angles.

### b. Let \( M \in CB \) such that \( CM = CA \). Draw \( CE \) perpendicular to the angle bisector \( ACN \).

1. Since \( CM = CA \), triangle \( CAM \) is isosceles with \( AM = AC \).
2. From the properties of triangle bisectors and isosceles triangles, you can show that \( ME \perp AN \) because it also relates to the right triangle properties and corresponding angles.

### c. Let \( CE \) intersect \( BD \) at \( O \) and \( AM \) at \( F \), also \( BD \) intersects \( AM \) at \( I \). Prove that \( IE = AO \).

1. Since \( O \) and \( A \) are points on lines intersecting \( BD \) and \( CE \) with the given conditions, you can apply triangle congruence.
2. Specifically, focus on triangles formed by \( I, E, O, \) and \( A \) and prove congruences using the angle bisector theorem and properties of similar triangles.

### Conclusion

Using properties of triangles, angles, and congruences will lead to the required proofs as outlined in each part of the problem. Each step follows logically from the previous, leveraging the relationships between the points and the angles defined.