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IJ = AH wanted, parallel , starting with northocenter of an acute triangle

Source: Switzerland - Swiss MO 2016 p8

July 15, 2020
geometryorthocenterequal segments

Problem Statement

Let ABCABC be an acute-angled triangle with height intersection HH. Let GG be the intersection of parallel of ABAB through HH with the parallel of AHAH through BB. Let II be the point on the line GHGH, so that ACAC bisects segment HIHI. Let JJ be the second intersection of ACAC and the circumcircle of the triangle CGICGI. Show that IJ=AHIJ = AH
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