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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click on the figure below to enlarge it.
Since ∡ CEA = ∡ CDA , A,B,C,D,E are concylicLet α be ∡CDE, then ∡CAE = α ; ∡ EAD = ∡ECD = 45 - α ; ∡FCD = ∡FCE - ∡ECD = 45 - (45 - α) = α =∡CDE So FC//DEFurther ∡ACF = 45 - ∡FCD = 45 - αLet L be the length of the square,By sine law on △CDE, L/sin(135) = 1/sin(45 - α)By sine law on △ACF(sqrt2)L / sin(135) = x/sin(45 - α)Hence, x = sqrt2
Let AD=a, CE=r. Since ADEC is cyclic quadrilateral, by Ptolemy theorem, AC×DE + AD×CE = AE×CD√2 a + ar = a(x+r)x = √2
a geometric solution: http://bleaug.free.fr/gogeometry/934.pngBy construction, ABCDE are co-cyclic. Let G be co-cyclic such that AC crosses CG in F. By construction AFG=EFC=45°AD chord in ABCDEG circle implies ACD=AED=45°. Similarly, CD chord implies CAD=CGD=45°Hence ED // CG and GD // AE, therefore EDGF is a parallelogram.AF = GF.√2 = ED.√2 = √2
Very nice solution!
<ACF=<DCE=45-<FCD, and <CDE=<CAF due to cyclic quadrilateral. Therefore X/CF=1/CE, X=CF/CE=sqrt2