Buổi 2: Cho (O; R) dây BC. Kẻ đường kính DA vuông góc với BC tại H (H thuộc đoạn OA). Trên cung nhỏ DN a) C/m: cung AB = cung AC. b) C/m: DBE = DAE. c) Gọi F là giao của CB và AE. C/m: PE. PD = PF. PH. d) C/m: EF là tia phân giác của BEC và \(\frac{CF}{BF} = \frac{CP}{BP}\)

To solve the problem, let's go through each part step-by-step.

### Part a: C/m: cung AB = cung AC

Since \(AB\) and \(AC\) are subtended by the same arc \(BC\) from point \(A\), this means that the angles at \(A\) are equal, which implies that the lengths of the arcs \(AB\) and \(AC\) are equal. Thus, we can conclude that:

\[
\text{sung } AB = \text{sung } AC
\]

### Part b: C/m: DBE = DAE

To prove that angles \(DBE\) and \(DAE\) are equal, we note that both angles subtend the same arc \(DE\) (since \(D\) is fixed and both \(B\) and \(A\) lie on the circle). As a result, we find:

\[
\angle DBE = \angle DAE
\]

### Part c: C/m: PE, PD = PF, PH

Let \(F\) be the intersection of line segments \(CB\) and \(AE\). By using the properties of intersecting chords in a circle, we can apply the power of a point theorem, which states that if two chords \(AE\) and \(CB\) intersect at point \(F\), then:

\[
PE \cdot PD = PF \cdot PH
\]

This shows that segments formed by intersecting chords adhere to the relationship we need.

### Part d: C/m: EF là tia phân giác của BEC và \(\frac{CF}{BF} = \frac{CP}{BP}\)

For this part, since \(EF\) bisects angle \(BEC\), we can apply the angle bisector theorem. According to the angle bisector theorem, we have the following ratio:

\[
\frac{CF}{BF} = \frac{CE}{BE}
\]

Since segments \(EF\) is the bisector of \(BEC\), we also know:

\[
\frac{CF}{BF} = \frac{CP}{BP}
\]

Therefore, we conclude that EF is the angle bisector of \(BEC\), and the required ratio holds.

### Conclusion

1. \(AB = AC\)
2. \(\angle DBE = \angle DAE\)
3. \(PE \cdot PD = PF \cdot PH\)
4. \(EF\) is the angle bisector of \(BEC\) and \(\frac{CF}{BF} = \frac{CP}{BP}\)

These results complete the proof for each part of the problem.