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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 736 details.
http://img826.imageshack.us/img826/5130/problem736.pngQuadrilateral DBEH is concyclicIn right triangles AHB and BHC we have BD.BA=BH^2=BE.BC => quadrilateral ADEC is concyclic Let O is the center of cycle ADEC. (See sketch)In circle DBEH, ∠ (HBE)= ∠ (HDE)=33 => ∠ (EDC)=33-21=12 Angles at the center of circle ADEC ∠ (AOE)=2 *57=114 and ∠ (EOC)=2*12=24So ∠ (AEB)=x=1/2*(114+24)=69
Triangles BDH and BHA are similarSo are trangles BEH and BHCTherefore BD.BA = BH^2 = BE.BCSo D,A,C,E are concyclic<BDE = <ECA = 57 deg<EDC = 90 - 57 -21 = 12 deg<EAC = <EDC = 12 degHence x = 57 deg + <EAC = 57 + 12 = 68 deg
Typo 57 + 12 = 69
The solution is uploaded to the following link:https://docs.google.com/open?id=0B6XXCq92fLJJeHNTbEVCdnk1LU0
BDHE is concyclic hence < BDE = < BHE = < BCH = 57 so ADEC is also concyclicHence < CDE = 12 = < EAC. Therefore x =12 +57 = 69Sumith PeirisMoratuwaSri Lanka