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 811.
Consider ∆DAE and ∆EAF. They are both isosceles and ∠DAE=∠EAF. Therefore, ∆DAE ~ ∆EAF. So AD / AE = AE / AFBut AD = CB + AB = 3*AB = 3*AEThusAB / AF = AE / AF = AD / AE = 3Hence, AF = AB / 3.
Triangles AEF and ADE are isosceles.Note that ∠EAF=∠EFA=∠EAD=∠AEDSo triangle AED similar to ADE ( case AA)And AF/AE=AE/AD=1/3 => AF=1/3.AE=1/3.AB
Consider the inversion transform T = I(A, AB^2). T(Circle A) = Circle AT(Line CAB) = Line CABNow consider circle E. T(circle E) would pass through the intersection points of circle A and circle E. And since T(circle E) should be a line that does not pass through A, so T(circle E) = the common chord of circle A and circle E. Since the common chord of circle A and circle E is exactly the perpendicular bisector of AE, which should pass through point D. T(circle E ∩ line CAB) = T(circle E) ∩ line CABThus, T(F) = D. By the definition of inversion, AF * AD = AB^2But AD = 3*AB, hence, AF = AB / 3.