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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view the complete problem 989
http://s25.postimg.org/vsrf9vrnj/Prob_989.pngLet M, L, N, O are points shown on the sketch1. We have MG⊥GH and NH⊥GHSince LG=LB=LH=> L is the midpoint of GH∠(GBH)= ½(∠GMB+∠BNH)= 90And ∠ (MLN)=1/2(∠GLB)+ ∠BLH)=90In right triangle MLN , BL^2=BM.BNIn right triangle ADC , DB^2=BA.BC= 4. BM.BN=> L is the midpoint of BDAnd DGBH is a parallelogram with right angle ∠GBH => DGBH is a rectangle2. Connect AG and CHWe have ∠ (AGB=∠ (BGD)= 90 => A,G,D are collinearTriangles AMG and AOD are isosceles with common angle ∠GAMSo ∠ (AGM)= ∠ (ADO) => OD//MG => OD ⊥ EFSo D is the midpoint of arc EF
I have a comment on the second part.DO || HN, and they are perpendicular to EF.So DO bisects both chord EF and arc EF.
Thank you for your comment on my solution.Peter
Because <GAC=<FHC, AGHC is cyclic. Let AG and HC meet at D'. <D'AC+<D'CA=90, so D' must be on big circle. Furthermore, because D'G*D'A=D'H*D'C from cyclic property of AGHC, D' must also be on radical axis of 2 small circles, which is their common tangent. Therefore, D' coincides with D. It is easy to see that GDHB has 3 right angles from Thales theorem in each circle and must therefore be rectangle.<GAC=<FHC=arcDF/2+arcFC/2=arcDE+arcFC/2. This means that arcDF=arcDE so D is midpoint of arcEF.