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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 671.
http://img838.imageshack.us/img838/3824/problem671.pngDraw triangle BDE as image of triangle BDA ( see picture)Note that BE=CD and (DCB)=(EBC)=4. alphaAnd DBCE is an isosceles trapezoid (BDE)=(BDA)= 180-7.alphaIn triangle ABD , x+180-7.alpha+3.alpha= 180 >> x=4. alphaIn triangle ABC , 7.alpha+4.alpha+4.alpha=180 >> alpha=10 and x=40Peter Tran
Okay, I have a really simple solution.At first the problem seems to be awful but it can be taken down faster.Let P be a point on BC so that DB = DP, hence <CDP = 3a and then ∆CDP = ∆ABD and x = 4a, hence a = 10° whereat x = 40°.
Very good simple solution.
Find E on DC such that < CBE = 3@Let BD and BE extended meet the circle ABC at X and Y.AX = CY so ACYX is an isoceles trapezoid.Now < AEB = 7@ = < ABEHence AB = c = AE = CDSo AD = CE = b-cFurther < XAC = YCASo tr.s AXD and CYE are congruent SAS. < BDC = < BEA = 7@Hence 7@ + 7@ + 4@ = 180 and therefore @ = 10 and x = 40Sumith PeirisMoratuwaSri Lanka