Question Number 189054 by mathlove last updated on 11/Mar/23
Answered by som(math1967) last updated on 11/Mar/23
$$\angle{EBD}=\angle{HBC}\:,\angle{ABC}=\angle{EBF} \\ $$$$\angle{HBF}=\angle{ABD} \\ $$$$\angle{EBD}+\angle{EBF}+\angle{HBF}+\angle{HBC} \\ $$$$+\angle{ABC}+\angle{ABD}=\mathrm{360} \\ $$$$\mathrm{2}\left(\angle{EBD}+\angle{HBF}+\angle{ABC}\right)=\mathrm{360} \\ $$$$\angle{EBD}+\angle{HBF}+\angle{ABC}=\mathrm{180} \\ $$$$\angle{EBD}+\mathrm{40}+{n}+\angle{HBF}+{k}+\mathrm{55} \\ $$$$+\angle{ABC}+{m}+\mathrm{90}=\mathrm{3}×\mathrm{180} \\ $$$${m}+{n}+{k}+\mathrm{185}=\mathrm{540}−\mathrm{180} \\ $$$${m}+{n}+{k}=\mathrm{360}−\mathrm{185}=\mathrm{175} \\ $$