Heights AA1 and CC1 of an acute-angled triangle The heights AA1 and CC1 of an acute-angled triangle ABC meet at point E. Prove that angles CC1A1 and CAA1 are equal.
Let us prove that right-angled triangles AEC1 and CEA1 are similar.
The angles AEC1 and CEA1 are equal as the vertical angles at the intersection of straight lines AA1 and CC1, then the triangles are similar in acute angle.
AE / CE = EC1 / EA1.
CE * EC1 = AE * EA1.
Let’s rewrite the proportion in the form: EC1 / AE = EA1 / CE, which proves the similarity of triangles A1C1E and ACE, then the angle CC1A1 and CAA1 are equal. Q.E.D.
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