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

May 31, 2020 | Education

| 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

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