tag:blogger.com,1999:blog-6933544261975483399.post446783907984425966..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1009: Tangent Circles, Common External Tangent, Common Internal Tangent, Arithmetic MeanAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-71519616701056298632018-08-24T11:52:59.668-07:002018-08-24T11:52:59.668-07:00I forgot the add the final piece:
DC=EF because ...I forgot the add the final piece:<br /><br />DC=EF because AB=DC and therefore DC=2DF=2AE then<br /><br />2EF=AD+BC<br />EF=(AD+BC)/2Arifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84256403750947949572018-08-23T11:28:42.936-07:002018-08-23T11:28:42.936-07:00Join the points CT and DT.
Also join the points AT...Join the points CT and DT.<br />Also join the points AT and BT.<br /><br />We have <br /><br />DF=FT=FC=DC/2<br /><br />From the similarity between triangles DFT and DTA we get <br /><br />DT/AD=DC/2DT<br />(1.)2DT^2=(AD)(DC)<br /><br />And from the similarity between triangles CFT and CTB we get <br /><br />CT/BC=DC/2CT<br />(2.)2CT^2=(BC)(DC)<br /><br />Adding equation 1 and 2 we get <br /><br />2(DT^2+CT^2)=(AD+BC)DC<br /><br />Because DTC is a right triangle we have <br /><br />2DC^2=(AD+BC)DC<br />2DC=(AD+BC)Arifhttps://www.blogger.com/profile/16774426807135540272noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21608975950422929672014-04-25T16:55:54.445-07:002014-04-25T16:55:54.445-07:00Since EA=ET=EB, FC=FT=FD,
Also AD//BC (ABCD is tr...Since EA=ET=EB, FC=FT=FD, <br />Also AD//BC (ABCD is trapezium), <br />Thus EF=(AD+BC)/2. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com