MathDB

P18

Part of 2017 Sharygin Geometry Olympiad

Problems(1)

angle relation with symmedian point and feet of altitude

Source: Sharygin Geometry Olympiad 2017 First Round P18 grades 9-11

7/22/2018
Let LL be the common point of the symmedians of triangle ABCABC, and BHBH be its altitude. It is known that ∠ALH=180o−2∠A\angle ALH = 180^o -2\angle A. Prove that ∠CLH=180o−2∠C\angle CLH = 180^o - 2\angle C.
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