tag:blogger.com,1999:blog-6500033298667240354.post7731683012520874903..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 2328foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6500033298667240354.post-11167682772874046382010-01-22T21:27:40.734-08:002010-01-22T21:27:40.734-08:00By similar triangles, one can prove that the commo...By similar triangles, one can prove that the common tangent will intersect with the x-axis at[-a(b+a)/(b-a),0] Further the length of the common tangent will be 2√(ab) which gives us 2C = ∏ab while A=∏a^2/2 & B =∏b^2/2 and thus A + B + 2C = ∏a^2/2 + ∏b^2/2 + ∏ab = (∏/2)(a+b)^2 -----(1) whereas D=∏((a+b)/2)^2/2 or 4D= (∏/2)(a+b)^2=A+B+2C by equation (1).<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com