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Category: Geometry

Question-201322

Question Number 201322 by cherokeesay last updated on 04/Dec/23 Answered by AST last updated on 05/Dec/23 $${Let}\:\angle{EDC}=\beta\Rightarrow\angle{ABD}=\mathrm{2}\beta−\mathrm{90} \\ $$$${EC}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cos}\beta…\left({i}\right) \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{BD}\boldsymbol{{D}}{Ccos}\left(\mathrm{2}\beta\right)…\left({ii}\right) \\…