Let's go through the problem step by step:

### a) Chứng minh △AEF cân.

To prove that triangle \( AEF \) is isosceles, we need to demonstrate that \( AE = AF \).

1. Since triangle \( ABC \) is isosceles with \( AB = AC \), we know that angles \( ABC \) and \( ACB \) are equal.
2. Points \( E \) and \( F \) lie on the extended lines of \( BC \) and \( CB \), respectively, such that \( BE = CF \).
3. By the properties of isosceles triangles and the equal lengths \( BE \) and \( CF \), it follows that angles \( AEF \) and \( AFE \) must be equal, leading to \( AE = AF \).

Thus, triangle \( AEF \) is isosceles.

### b) Vẽ \( BH \) vuông góc với đường \( AE \). Vẽ \( CK \) vuông góc với đường \( AF \). Chứng minh \( \angle EBH = \angle FCK \).

1. By construction, \( BH \) is perpendicular to \( AE \) and \( CK \) is perpendicular to \( AF \).
2. Since \( \triangle AEF \) is isosceles, angles \( \angle AEF \) and \( \angle AFE \) are equal.
3. Then \( \angle EBH \) and \( \angle FCK \) are alternate interior angles formed by the transversal line \( AE \) and \( AF \) intersecting the lines \( BH \) and \( CK \), confirming that \( \angle EBH = \angle FCK \).

### c) Các đường thẳng \( HB \) và \( KC \) cắt nhau tại \( I \). Tam giác \( IHK \) là tam giác gì? Tại sao?

1. Since lines \( HB \) and \( KC \) are drawn perpendicularly to \( AE \) and \( AF \) respectively, these lines imply certain angles at point \( I \).
2. The angles formed will be right angles (\( 90^\circ \)) due to the perpendicularity.
3. Therefore, triangle \( IHK \) will be a right triangle.

### d) Khi \( \angle BAC = 60^\circ \) và \( BE = CF = BC \), tính số đo các góc của tam giác \( AEF \) và xác định dạng.

1. Since \( AEF \) is isosceles, let's denote the angles as \( x \) at vertices \( E \) and \( F \) and \( 60^\circ \) at vertex \( A \).
2. Therefore, we have:
\[
x + x + 60^\circ = 180^\circ \Rightarrow 2x = 120^\circ \Rightarrow x = 60^\circ
\]
3. This implies that triangle \( AEF \) has angles \( 60^\circ, 60^\circ, 60^\circ \), thus making it an equilateral triangle.

### Kết luận:
- \( \triangle AEF \) là tam giác đều với tất cả các góc đều bằng \( 60^\circ \).