Geometry Problem
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Level: High School, SAT Prep, College geometry
Sunday, December 26, 2010
Problem 556 Quadrilateral, Diagonal, Angles, Auxiliary Lines.
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Bisect Angle BCD with CE (E is on AD), connect BE; bisect Angle EBD with BF (F is on AC), connect EF.
ReplyDeleteTriangles 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!
ReplyDeleteDraw 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 deg
Hence x + 39 deg = 69 deg,
x = 30 deg
Nice problem
ReplyDeletetwo nice solutions