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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 773.
http://img59.imageshack.us/img59/328/problem773.pngDraw new lines per attached sketchWe have ∠BDA=1/2*∠BQA= ∠BQOAnd ∠BCA=1/2*∠BOA =∠BOQ ∆BOQ similar to ∆BCD …. ( case SAS)∆ BCD is the image ∆BOQ in the spiral similarity geometry transformation with following details. center = point B , angle of rotation θ= ∠OBC=∠QBD , ratio of similarity= BC/BO=BD/BQSince M and E are midpoints of OQ and CD , so E is the image of M in this geometry transformation. So we have ∆BME similar to ∆BOC => MB=ME and E is the point of intersection of CD and circle M
Please draw a figure Let F be diametrically opposite to A w.r.t. cirle (M). Join EF. So ∠AEF = 90°, E is the foot of ⊥ from F on CD.Let O', M', Q' be the feet of the ⊥s from O, M, Q on CD.So CO' = O'A, AM' = M'E, AQ' = Q'DJoin OF, QF.OAQF is a parallelogram since diagonals OQ, AF bisect each other.OA = FQ and OF = AQSo O'A = EQ' and O'E = AQ'Now CE = CA + AE = 2O'A + AE = 2EQ' + AE = EQ' + (EQ' + AE)= EQ' + AQ' = EQ' + Q'D = ED as required.
Ref: For figure of Problem 773, please go to the following link:http://s1092.photobucket.com/albums/i418/vprasad_nalluri/?action=view¤t=GoGeometryProblem773.jpg
Let OC=r and QD=RThis is true, if and only ifEA*EC=ED*EA orEO^2-r^2=R^2-EQ^2 orR^2+r^2=EO^2+EQ^2 orAO^2+AQ^2=EO^2+EQ^2 or2(OQ^2)+(AM^2)/2=2(OQ^2)+(EM^2)2 orAM=EM.But this is true
Let X,Y, and Z be the feet of the perpendiculars from O,M, and Q to CD. Then XA=(1/2)AC and AZ=(1/2)AD, so XZ=(1/2)CD, so XY=YZ=(1/4)CD, so XY=XA+AY=(1/2)AC+(1/2)AE=(1/2)CE=(1/4)CD, so CE=(1/2)CD.
(O ,r) ,(Q,R) The two circlesEA.EC= EO^2 - r^2, EA.ΕD=R^2-EQ^2 EA.EC- EA.ΕD= EO^2+ EQ^2-( r^2+ R^2)=2ΕΜ^2+2ΟΜ^2-2ΑΜ^2-2ΟΜ^2=0(by Theorem median in triangles ΟΕQ,OAQ, and because ΑΜ=ΕΜ ).So, EA.EC- EA.ΕD=0 ,EA.EC=EA.ΕD, EC=ED