The heights BB1 and CC1 of the acute-angled triangle ABC intersect at point E. Prove that the angles BB1C1 and BCC1 are equal

Draw a segment B1C1 and consider the triangles EB1C and EC1B.
∠C1EB = ∠B1EC (since they are vertical).
∠EB1C = ∠EC1B = 90 ° (since BB1 ​​and CC1 are heights).
According to the first sign of the similarity of triangles, the considered triangles are similar.
Hence:
EB1 / EC1 = EC / EB
Consider the triangles EC1B1 and ECB
∠BEC = ∠B1EC1 (since they are vertical).
As we found out earlier:
EB1 / EC1 = EC / EB
We multiply the left and right sides of the equality by EC1, we get:
EB1 = EC1 * EC / EB
Divide the left and right parts by EC, we get:
EB1 / EC = EC1 / EB
It turns out that according to the second sign of the similarity of triangles, the triangles EC1B1 and ECB are similar.
Therefore, by definition, the angles BB1C1 and BCC1 are equal

Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.