MathDB
point symmetric to circumcneter of (HOA' ) lies on midline

Source: All-Russian MO 2009 Regional 11. 4

August 27, 2024
geometrymidline

Problem Statement

In an acute non-isosceles triangle ABCABC, the altitude AAAA' is drawn and point HH is the intersection point of the altitudes and and OO is the center of the circumscribed circle. Prove that the point symmetric to the circumcenter of triangle HOAHOA' wrt straight line HOHO, lies on a midline of triangle ABCABC.