Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1031.
See sketch for location of S, L, NWe have SE^2=SA.SB and SF^2=SC.SDSB/SA=SC/SD1. Calculate p^2= (SE.SD)^2= SA.SB.SD^2And p1^2=(SA.SF)^2= SC.SD.SA^2(p/p1)^2= (SB/SA).(SD/SC)= 1So p=p1 => quadrilateral AFED is cyclicSimilarly quadrilateral BFEC is cyclic2. We have ∠(FAE)= ∠(FDE)= ∠(BFC)= ∠ BEC) => AE//FC and FD//BEAnd FNEL is a parallelogramWe also have NE.NA=NF.ND => N will be on radical line GH of circles O and QSimilarly L will be on GHM is the intersection of 2 diagonals of a parallelogram FNEL so M is the midpoint of EF.
To Peter Problem 1031Sketch link?
below is the sketch of problem 1031http://s25.postimg.org/xlwtf6fjz/pro_1031.pngPeter Tran