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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 839.
∠EBA = 180-∠ABC = 180-∠ADC = ∠ADF Triangle EDA ~ Triangle ADF (AA)Hence EB/AB = DF/AD DF = 15/4
Let G be a point such that EAGC is rectangle. Then CD=4, DG=3. ∵ A,F,G,C concyclic∴ DF×DC=DA×DG⇒ 4x=15, x=15/4
Aria paralelogramului ABCD=BC.AE=AF.DC=>AF=BC.AE/DC=5√7/4AE^2 = 4^2 −3^2 = 7,AD^2 = AF^2 +FD^2 ,x^2 = 5^2 −(5√7/4)^2 = 5^2(1-7/16)=25.9/16=>x=15/4