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Geometry ProblemClick the figure below to see the complete problem 515 about Triangle, Double Angle, Altitude, Measure.
Locate E on CD s.t. DE=d. Now triangles BAD & BED are congruent with BE=c and /_BED=2α which makes /_EBC=2α-α=α or BE=EC=c. Hence, e=DE+EC=d+cAjit
extend CA to G, AG = AB => ▲GBC isoceles =>BD median
Extend AC to the left and locate point E such that AE=c Triangle ABE isosceles with angle(AEB)=alphaso triangle EBC isosceles and BD is the median of isosceles tri. EBCwe have DC=DE=DA+AE=d+cPeter Tran
Let point E on line AC, and BE=BC.Angle BEA=angle BCA=alpha.AB=AE, since angle ABE=alpha (2xalpha - alpha).Point D=midpoint of EC---> EA+AD=DC (c+d=e)