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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 857.
CA ,perpendicular to AB. Then, angle ACE=60 ,and angle ECB=45. Because, angle EBC=45 ,EC=EB=ED=DA. So , 2x=30 ,x=15 degreeshttp://img571.imageshack.us/img571/3688/p857trianglemedianmidpo.gif
Reflect C along AB to a point E.Angle(EBC)=2*Angle(ABC)=2*(45)=90By reflection, Triangle(AEC) is equiliteral andsince AD=DC, Angle(EDC)=90.So, E,D,B,C are concyclic.x = Angle(ABC)-Angle(DBC) = 45 - Angle(DEC) = 45 - (1/2)*(Angle(AEC)) = 45-30 = 15.
http://img42.imageshack.us/img42/6031/problem857.pngDraw CE ⊥ABWe have ∠ (ACE)= 60 and ED is a median of right triangle AECSo ED=DA=DCWe have ∠ (ECB)=105-60=45 => EC=EB=ED∠ (AED)=30=2.x =>x=15
This kind of problems usually are solvable by placing the circumcenter.- Let O be the circumcenter of (ABC).- AD=OD=CD, so tr DOB is isosceles.- tr OBC is equilateral.- ODCB is a deltoid, so BD is perp to OC and <OBD=30°.- <OBA=<OAB=15°.Finally <x=15°.