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2KF+BC=BH +HC, orthocenter, incenter , projection related

Source: 2011 Belarus TST 7.1

June 14, 2020
geometryincenterorthocenterperpendicular

Problem Statement

In an acute-angled triangle ABCABC, the orthocenter is HH. IHI_H is the incenter of △BHC\vartriangle BHC. The bisector of ∠BAC\angle BAC intersects the perpendicular from IHI_H to the side BCBC at point KK. Let FF be the foot of the perpendicular from KK to ABAB. Prove that 2KF+BC=BH+HC2KF+BC=BH +HC
A. Voidelevich
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