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It seems like you've uploaded a math problem involving circles and geometric relationships. Let me summarize and break down the problem based on the image you provided:

The problem is related to a circle with a specific geometric configuration:
- There is a circle (O) with radius BC, and a point A on the circle.
- Points A and B lie on the circle, and the length of AC is greater than AB.
- There is another circle (O') passing through point A, and this circle intersects BC at points C and B.
- BF is a line intersecting circle (O) at point E. BE is the perpendicular bisector of segment HN, and you are tasked with proving the following:
  a) Points C, E, F, and H lie on the same circle.
  b) \( BF \cdot BE + CF \cdot CA = BC^2 \).
  c) Points A, N, and E are collinear.

Now, let's start with part a and begin proving the geometric relationships. Would you like me to continue with this step-by-step approach and explain the reasoning for each part, or do you have specific sections you'd like to focus on?

Cho đường tròn (O) đường kính BC và điểm A nằm trên đường tròn sao cho AC > AB > 0. Đường tròn (O') đi qua A, B cắt các cạnh BC, CA tương ứng tại các điểm thứ hai H và F. Tia BF cắt đường tròn (O) tại điểm thứ hai E. Về điểm N, E thuộc đường thẳng HN. Chứng minh rằng:
 a) Bốn điểm H, C, E, F cùng thuộc một đường tròn;
 b) BF . BE + CF . CA = BC^2;
 c) Ba điểm A, N, E thẳng hàng.

This problem involves advanced geometric principles including the Power of a Point theorem, properties of cyclic quadrilaterals, and collinearity of points. It is suitable for students in grades 10-12 and focuses on proving relationships within circles and geometric figures.

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