Let's go through each part of the problem step by step.

### a) Chứng minh: ΔABD = ΔEBD.

To prove that triangles \(\Delta ABD\) and \(\Delta EBD\) are congruent, we need to verify:

1. \(BD\) is common in both triangles.
2. \(\angle ABD = \angle EBD\) because \(BD\) is the angle bisector.
3. \(\angle ADB = \angle EDB = 90^\circ\) since \(DE\) is perpendicular to \(BC\).

Using the AAS (Angle-Angle-Side) criterion, \(\Delta ABD = \Delta EBD\).

### b) Chứng minh: BD là đường trung trực của đoạn thẳng AE.

Since \(\Delta ABD = \Delta EBD\), we have:
- \(AD = DE\) (corresponding parts of congruent triangles).
- \(\angle ADB = \angle EDB = 90^\circ\).

Thus, \(BD\) is the perpendicular bisector of segment \(AE\).

### c) Gọi K là giao điểm của BA và ED. Chứng minh: ΔBKC đều.

To prove \(\Delta BKC\) is equilateral, note:
- \(\angle ABC = 60^\circ\).
- \(ED \perp BC\), so \(\angle BKC = 90^\circ - 30^\circ = 60^\circ\).
- \(\angle BCK = 60^\circ\).

All angles in \(\Delta BKC\) are \(60^\circ\), making it equilateral.

### d) Gọi I là trung điểm của CK. Chứng minh: B, D, I thẳng hàng.

Since \(\Delta BKC\) is equilateral:
- \(BI\) is a median and also an altitude of \(\Delta BKC\).
- As \(BD\) is the bisector perpendicular to \(AE\), and parallel lines (MEDIANS) are concurrent, points \(B\), \(D\), and \(I\) align on a straight line.

Thus, \(B\), \(D\), \(I\) are collinear.