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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 enlarge the problem 942.
http://img22.imageshack.us/img22/3034/goav.pngDraw points L, M, F’, I’ per sketch.We have AL=CM= p-aSince F is the midpoint of arc AC so B, I, F, E are collinearAnd D and F are midpoints of LM and IEDF=MF’= FI’- I’M= ½( 20+7)- 7 = 6.5
ArcAF=ArcAC so BF bisect angleABC and hence BIFE collinear Let BF cut AC at P ; incircle and ex circle cut AC at I' and E' Note Triangle II'P ~ PDF ~ PEE'Further D is the midpoint of AC hence the midpoint of I'E'By solving similar triangle ratios it is arrived that L = 13/2
Let IJ be the in-radius shown in the graph, and EK be that ex-radius shown. Observe that JDK are collinear with mid-point D. Also, IFE are collinear, which is the angle bisector. Thus, DF = 1/2 (20 - 7) = 13/2