Geometry Problem

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Level: High School, SAT Prep, College geometry

## Sunday, December 26, 2010

### Problem 556 Quadrilateral, Diagonal, Angles, Auxiliary Lines.

Labels:
angle,
auxiliary line,
diagonal,
quadrilateral

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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