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Geometry ProblemClick the figure below to see the complete problem 556.ZoomLevel: High School, SAT Prep, College geometry
Bisect Angle BCD with CE (E is on AD), connect BE; bisect Angle EBD with BF (F is on AC), connect EF. Triangles BCE & DCE are congruent, Angle CBE=78 Degrees. Since Angles CBE and BCD are bisected, Angle EBF=Angle CBF=39 Degrees. Angle BCE=Angle DCE=42 Degrees. So AC bisects Angle BCE, Angle ACB=Angle ACE=21 Degrees. Thus, Triangle BCE's incenter is F (EF bisects Angle BEC, Angle BEF=Angle CEF=30 Degrees. ABEF is concyclic (two 39 Degree angles subtended on EF), therefore x=30 Degrees.
A nice problem indeed!Draw an arc centre at C with radius CB cutting AD at E.CE = CD => AE = CE CB = CE, Triangle BCE is equilateral Follows AE = BE, <BAE (= <ABE) = 69 degHence x + 39 deg = 69 deg, x = 30 deg
Nice problemtwo nice solutions