tag:blogger.com,1999:blog-6933544261975483399.post1641804230086651113..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 313. Circle, Chord, Tangent, Perpendicular, Geometric MeanAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-67868193636719340092018-02-02T13:11:31.298-08:002018-02-02T13:11:31.298-08:00Connect EC and ED.
We see that triangle EBC and ...Connect EC and ED. <br /><br />We see that triangle EBC and DFE are similar therefore : <br /><br />a/EC=c/DE<br />(1.) a/c=EC/DE<br /><br />and triangle EAD and CFE are similar therefore <br /><br />c/EC=b/DE<br />(2.) c/b=EC/DE <br /><br /><br />From equations 1 and 2 we have <br /><br />c/b=a/c<br /><br />c^2=abArifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74260384888433938092009-07-03T21:43:42.428-07:002009-07-03T21:43:42.428-07:00Join C & D to E. By the tangent-chord theorem ...Join C & D to E. By the tangent-chord theorem angle BEC = angle CDE and angle AED = angle ECF. This makes triangles EAD & EFC similar since angle EAD=angle EFC=90. Likewise, triangles CBE & EFD are similar. This gives us: b=(EA/CF)c and a=(BE/DF)c as also EA/CF=DE/CE while BE/DF=CE/DE<br />We can therefore say that a*b =(EA/CF)c*(BE/DF)c <br />or ab =c^2*(EA/CF)*(BE/DF)=c^2*(DE/CE)*(CE/DE)=c^2. QEDAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com