Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 676.
http://img18.imageshack.us/img18/5852/problem676.pngConnect DB, DE and note that AB perpendicular to BC, DB per. To AC, DE per. To AB and EF per. To ACWe have DB^2=AD.DC= a.b ( relations in right triangle)So DB=SQRT(a.b)Triangle AEF similar to tri. ABD and triangle AED similar to tri. ABCSo we have x/DB=AE/AB=AD/AC=a/(a+b)And x=a.DB/(a+b)=a.SQRT(a.b)/(a+b)Peter Tran
Nice short solution!I needed more work :-(
x /√(ab) = x / BD = AE / AB = AD / AC = a /(a+b)(∵ BD² = AD.DC; EF ∥ BD; DE ∥ CB)
We have CB _|_ to AB => CB^2 = CD.CA=> CB = Sqrt(b.(a+b))and AB = Sqrt(a.(a+b))Connect DE and consider the right triangle ADE. ADE is similar to ACB=> AE/AB = AD/AC=> AE = a.Sqrt(a.(a+b))/(a+b)=> AE = a.Sqrt(a)/Sqrt(a+b)Now consider the right triangle AFE which is similar to ABC=> EF/CB = AE/AC=> EF = x = [a.Sqrt(a)/Sqrt(a+b)].Sqrt(b.(a+b))/(a+b)=> x = a.Sqrt(ab)/(a+b)