The image you've uploaded is in Vietnamese and seems to be a geometric proof. It describes a right triangle $\triangle ABC$ with specific geometric properties. Here's a breakdown:

1. Given: $\triangle ABC$ is a right triangle with the right angle at $A$.
2. Three altitudes are drawn: $AD$, $BE$, and $CF$.
3. The altitudes intersect at point $O$ (with $D \in BC$, $E \in AC$, and $F \in AB$).
4. The goal is to prove the equation: $BO \cdot BE + CO \cdot CF = BC^2$.

Let me analyze the geometry and set up the proof for you. Would you like me to proceed with a detailed explanation, including the geometric steps involved?

If you would like further clarification, feel free to ask!

Here are five related questions you might find interesting:

1. What is the general formula for the area of a triangle using its altitudes?
2. How do the lengths of the altitudes relate to the sides of a right triangle?
3. What is the significance of the orthocenter in a triangle?
4. How does the intersecting of altitudes help in geometric proofs like this?
5. What properties of right triangles can simplify this kind of geometric proof?

Tip: When dealing with geometric proofs, it's helpful to draw auxiliary lines like altitudes or medians, as they often reveal symmetries or relationships that can simplify the solution process.