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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 836.
Using Pythagoras Theorem, EB=3, EC=√65, AD=BC=√65-3Let G be a point such that EAGC is a rectangle. Then CD=5, DG=3. Since AFGC concyclic, AD×DG=FD×DCFD=3/5×(√65-3)Using Pythagoras Theorem again, in ΔADF, AF=4/5×(√65-3)Now since AECF concyclic, by Ptolemy Theorem, AC×EF = AE×CF + CE×AF9x = 4/5×[3√65+16] + 4/5×[65-3√65]9x = 4/5×81x = 36/5
SOLUTION by Michael TsourakakisBecause AB //CD, is ,FA perpendicular of AB.Let is ,CH perpendicular of AB. Then , AFCH is rectangle. So,the EHCA is, cyclic quadrilateral, additional, AECF is, cyclic quadrilateral, therefore , angle AHF=angle ACF=angleAEF,so, EHFA is, cyclic quadrilateral, , therefore , angle EHA=angleECA=angleCAD Then, the triangles AEF and ACD, are similar.So, x:9=4:5 therefore x=36:5see the image http://img26.imageshack.us/img26/8026/p836parallelogramperpen.gif
Let AC, BD intersect at O.The circle on AC as diameter passes through E, F. Its centre is O and diameter 2R = 9The angle θ subtended by its chord EF at C = ∠ECD = ∠EBAx = Length of the chord EF = 2Rsinθ (by sine rule) = 9(4/5) = 36/5