tag:blogger.com,1999:blog-6933544261975483399.post8288573297292730537..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1401: Right Triangle with three circles on the sides, Isosceles, Diameter, Center, Tangent, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-17256684547754378622018-12-03T10:51:35.482-08:002018-12-03T10:51:35.482-08:00It is easy to see that
e^2=(b/2)^2 -[(a-c)/2]^2=ac...It is easy to see that<br />e^2=(b/2)^2 -[(a-c)/2]^2=ac/2=Area of Tr. ABC <br /><br />Lets assume G' is midpoint of EF, if we draw perpendicular to AC from G', it will meet <br />AC at it's midpoint and its length will be (AE+CF)/2=AC/2. G' lies on circle c3, hence G' must coincide with G. <br />We have LG as median of Triangle ELF. <br />d^2=(EL^2+FL^2-2EG^2)/2 <br />where EL=EH and FL=FJ. <br />Lets assume EA extended meets c1 at point Q, then AQBD is a rectangle and AQ=BD <br />EH^2=EA.(EA+AQ)=EA^2+EA.BD .......(1) <br />Similarly FJ^2=FC^2+FC.BD ........(2) <br />Since Tr. EDF is right Triangle and G is midpoint of EF, <br />2EG^2=EA^2+FC^2 .......(3) <br />Using eq. (1),(2) and (3) we get d^2=(EA.BD+CF.BD)/2=AC.BD/2=Area of Tr.ABC <br /><br />Hence d=e Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com