MathDB
MC = MB + BH wanted, KH // AC, perpendiculars, external angle bisector

Source: 2019 RMM Shortlist G2

June 18, 2020
geometryparallelperpendicular bisectorangle bisector

Problem Statement

Let ABCABC be an acute-angled triangle. The line through CC perpendicular to ACAC meets the external angle bisector of ∠ABC\angle ABC at DD. Let HH be the foot of the perpendicular from DD onto BCBC. The point KK is chosen on ABAB so that KH∥ACKH \parallel AC. Let MM be the midpoint of AKAK. Prove that MC=MB+BHMC = MB + BH.
Giorgi Arabidze, Georgia,
Back to ProblemsView on AoPS