Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 673.
Draw the common tangent at A and join F to E. By the alternate segment theorem, /_FEA=/_CBA, both being equal to the angle made by BA with the common tangent. Thus, EF//BC which means f/CF=e/BE or f/(CF+f)=e/(BE+e).Further, by the power point theorem,(a-x)^2=BE(BE+e) and x^2=CF(CF+f) so [(a-x)/x]^2=[BE(BE+e)]/[(CF+f)CF] or (a-x)/x = e/f or x = (a*f)/(e+f)Ajit : email@example.com
http://img33.imageshack.us/img33/2368/problem673.pngConnect EF, O’D and AD ( see picture)Note that circle O is the image of circle O’ in the dilation transformation center A With dilation factor k= AO/AO’=AB/AE=AC/AFAnd BC//EF , O’C perpen. To BCAB is the angle bisector of (BAC)We have DC/DB=AC/AB=AF/AE=f/eReplace DC/DB=x/(a-x) in above expression we will get the result.Peter Tran
If we denote BD by y, by symmetry we havey = a.e/(e + f) and hence (following Tran)x/y = f/e = b/c = k = R/ρ where ρ is the radius of circle O'Follows ρ = cR/b. This answers"If the bisector of angle A of triangle ABC meets the opposite side BC at D, find the radius of the circle passing through A and touching BC at D."