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Acute-angled triangle with C=45

Source: Romanian NMO 2006, Grade 7, Problem 3

April 18, 2006
geometrycircumcircle

Problem Statement

In the acute-angle triangle ABCABC we have ACB=45\angle ACB = 45^\circ. The points A1A_1 and B1B_1 are the feet of the altitudes from AA and BB, and HH is the orthocenter of the triangle. We consider the points DD and EE on the segments AA1AA_1 and BCBC such that A1D=A1E=A1B1A_1D = A_1E = A_1B_1. Prove that a) A1B1=A1B2+A1C22A_1B_1 = \sqrt{ \frac{A_1B^2+A_1C^2}{2} }; b) CH=DECH=DE.