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Question Number 180273 by Shrinava last updated on 09/Nov/22

In triangle ABC with angles α , β , γ correspondently , Euler′s line interescts BC at point P. Ite′s put δ is angle between Euler′s line and BC (∠BPH). Then the following is true tan δ = ((2 cos β cos γ − cos α)/(sin (β − γ)))

$$\mathrm{In}\:\mathrm{triangle}\:\:\mathrm{ABC}\:\:\mathrm{with}\:\mathrm{angles}\:\:\alpha\:,\:\beta\:,\:\gamma \\ $$$$\mathrm{correspondently}\:,\:\mathrm{Euler}'\mathrm{s}\:\mathrm{line}\:\mathrm{interescts} \\ $$$$\mathrm{BC}\:\:\mathrm{at}\:\mathrm{point}\:\:\mathrm{P}.\:\mathrm{Ite}'\mathrm{s}\:\mathrm{put}\:\:\delta\:\:\mathrm{is}\:\mathrm{angle} \\ $$$$\mathrm{between}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{line}\:\mathrm{and}\:\:\mathrm{BC}\:\left(\angle\mathrm{BPH}\right). \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{tan}\:\delta\:=\:\frac{\mathrm{2}\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\gamma\:−\:\mathrm{cos}\:\alpha}{\mathrm{sin}\:\left(\beta\:−\:\gamma\right)} \\ $$

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