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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 688.
http://img403.imageshack.us/img403/6706/sfvdf.pngExtend AB from B till F so that < EFD = 30°. Since < CEF = 50°, then < CDF = 50° and DF ⊥ BC, whereat < BDF = 30°, and finally from the cyclic quadrilateral CDEF we have x + 30° = 50° and x = 20°.
Extend AB till F so that AF=AC. This way DCFE is an isosceles trapezoid(CD=CF=EF and <DCF=<CFE=80). <AED =80=<EBD+<BDE. So <BDE=20.
Did you both used AD=AE which isn't given or how can someone conclude EF=CD?
Reflection in the line BC. Let D→D'. ∵ ∠DBC = ∠D'BC = 60°∴ E, B, D' are collinear. ∵ ∠BDC = ∠BD'C = 80°∴ ∠D'CA = 80°, ∠D'CB = 40°Join DD'. Then BD = BD', ∴ ∠BDD' = 30°, ∠D'DC = 50°∵ ∠D'DC = ∠D'EC = 50°∴ C, D, E, D' are concyclic. ∴ ∠D'DE = ∠D'CE = 50°∴ x = ∠BDE = ∠D'DE − ∠BDD' = 20°
Reflection in the line CE. Let D→D'. Then ΔDCD' is equilateral. ∵ ∠DBC = ∠DD'C = 60°∴ B, D, C, D' are concyclic. ∵ ∠EBD = ∠DCD' = 60°∴ E, B, D' are collinear. In ΔACD', ∠AD'C = 100°∴ ∠ED'D = 40°∵ ED = ED'∴ ∠EDD' = 40°∵ ∠BDD' = ∠BDD' = 20°∴ x = ∠EBD = 20°
Find F on AB extended such that BC bisects < ACF. Then Tr. EFC is congruent with Tr. DFC and so EC = EF from which we can prove that Tr.s AFD and ACE are congruent. Hence AE = AD and so x = 20Sumith PeirisMoratuwaSri Lanka
Or simply EF = FC = DC and CDEF is concyclic so x+30 = 10+40 hence x= 20
Problem 688Draw the equilateral triangles DEK, then <EBD=<EKD=60.So EBKD is cyclic.Then <DEK=<DBK=60=<DBC.Therefore B,K,C are collinear. Is <EKD=60=2.30=2.<ECD,Then the point K is circumcenter of triangle DEC .Then x=<EDB=<EKB=10+10=20.MANOLOUDIS APOSTOLIS FROM GREECE