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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1025.
This problem may be solved in the next way:1.E be a point on AD such as DE = DC -> Angles: DEC = DCE = 2x2.Easy to see that DEC is congruent to BDC (SAS) -> EC = BC3.Angles: ECA = EAC = x = 3x - 2x -> EC = BC = AE4.Angles: BFE = 180 - BDE - BAD - EBD = 180 - 2x - 8x - (90 + 4x) = 90 -6x5.Let CG be perpendicular to EB thus EG = BG (Triangle EBC is isoceles)6.Let EF be perpendicular to AB 7.Triangle AFE ~= BCG ~= GCE (ASA) From simple look we get that in triangle BFE: EF = EG = BG -> AngleBFE = 30 = 90 - 6x (As we proved above)Finally x = 10.
Let AD meet BC at E and CD meet AB at F. ACEF is clearly cyclic from which we can easily show FE bisects < BED = 6x and FD bisects < ADB = 8x. So F is the excentre of Tr. BDE and so FB bisects the exterior < . So < ABC must be 8x and solving for x we get 8x + 2x + 5x + 3x = 180 from Tr. ABC and so x = 10Sumith PeirisMoratuwaSri Lanka