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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1207.
http://s22.postimg.org/nia3yt2fl/pro_1207.pngLet L, M, N , P are points as shown on the sketchObserve that AL= NC= half perimeter of triangle ABC- BCP is the midpoint of arc AC => M is the midpoint of AC and LNIn trapezoid ILND , MO is the mid-base => O is the midpoint of IDTriangle IOP similar to triangle IDE ..( case AA)Since O is the midpoint of ID so DE= 2 x OP= 2.R
Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.B, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.Consider triangle IED, O is midpoint of ID and H is midpoint of IE,Hence DE=2*OH, since OH=R, DE=2R.
Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.M is the midpoint of arc AMC. So OM bisects AC at right angles.We are done if we can show O is the midpoint of ID.Let X, Y, Z be the projections of I. O. D on AC respectively.It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.Hence DE = 2 OM = 2R.N Vijaya PrasadRajahmundry - INDIA.
Bring IK⊥AC,MN⊥AC, EL⊥AC(M is midpoint arc AC).Is <MIC=<ABC/2+<ACB/2 , <MCI=<MCA+<ACI=<ABC/2+<ACB/2.Therefore <MIC=<MCI.So MI=MC.But IC⊥CE.Then <MIC+<MEC=90 and <MCI+<MCE =90 therefore MI=MC=ME.So MO//EDIs IO=OD. Therefore ED=//2OM=2R.