To solve the geometric problems in the provided text, let's analyze each item separately.

### Bài 374

1. **Cho góc xOy khác góc bẹt**: Given that angle \( xOy \) is not a straight angle.

2. **Điều kiện**: Points \( A \) and \( B \) on line \( Ox \) such that \( OA < OB \), and points \( C \) and \( D \) on line \( Oy \) such that \( OC = OA \) and \( OD = OB \). Let \( E \) be the intersection point of \( AD \) and \( BC \).

a) **Chứng minh** \( \triangle EAB \cong \triangle ECD \):
- Since \( OA = OC \) and \( OB = OD \) and \( \angle EAB = \angle ECD \) (corresponding angles), we can use the criteria for triangle congruence (Side-Angle-Side or SAS).

b) **Chứng minh** \( OE \) là tia phân giác của góc \( xOy \):
- From the congruence of the triangles, if \( \triangle EAB \) and \( \triangle ECD \) are congruent, it follows that \( \angle OEB = \angle OED \), hence \( OE \) is the angle bisector.

### Bài 375

1. **Cho tam giác ABC có AB = AC**: Given triangle \( ABC \) is isosceles with \( AB = AC \).

a) **Chứng minh** \( \angle ABD = \angle ACD \):
- By properties of isosceles triangles, we have \( \angle ADB = \angle ADC \). Since \( D \) lies on line \( BC \), we can apply the properties of angles in a triangle.

b) **Lấy điểm E sao cho AE = AD**:
- If \( AE = AD \) and considering point \( F \) on line \( AB \) such that \( AF = AB \), it leads to proving \( EF = BD \) as well using triangle properties.

c) **Gọi H là trung điểm của FC**:
- Midpoint properties give us that \( AH \) is the angle bisector of \( \angle CAF \) due to symmetry and equal lengths from point \( A \) to the sides.

---

In summary, both problems revolve around properties of triangles, congruent triangles, and angle bisectors. You can apply congruence criteria and properties of isosceles triangles effectively to reach the required conclusions.