tag:blogger.com,1999:blog-6933544261975483399.post1076888418340036400..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1006: Tangent Circles, Common External Tangent, Bicentric TrapezoidAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-65907216649755847112015-09-17T10:58:57.962-07:002015-09-17T10:58:57.962-07:00Sorry typo.
Last line should be "with T as ...Sorry typo. <br /><br />Last line should be "with T as centre"Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10628227852325151092015-09-17T05:01:22.627-07:002015-09-17T05:01:22.627-07:00Extend AB, DC to meet at E. Then EBA and ECD are t...Extend AB, DC to meet at E. Then EBA and ECD are tangents drawn to the 2 circles from an external point E and are hence equal for each circle. <br /><br />So Tr.s EAD & EBC are isoceles and similar since they share the same angle <E. So BC//AD and < B = < D. Hence ABCD is a cyclic trapezoid with AB = CD. <br /><br />The circumcentre of ABCD must lie on OQ the perpendicular bisector of both AD and BC. Find O1 on OQ such that AO1 = BO1 and we can see easily thro congruence that these are also equal to DO1 and CO1. Hence O1 is the circumcentre <br /><br />Now draw the common internal tangent at T which is seen to bisect AB and CD so AT is perpendicular to BT and likewise DT to CT. Hence thro congruence we can show that the perpendiculars from T to ths 4 sides are all equal hence ABCD has an incircle with C as centre <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88161763416486518362014-06-09T05:40:39.401-07:002014-06-09T05:40:39.401-07:00Fie E,F ,G si H mijloacele segmntelor AB,BC,CDsi r...Fie E,F ,G si H mijloacele segmntelor AB,BC,CDsi respectiv AD=>EFGH patrat,(deoarece ABCD este un trapez isoscel si are diagonalele congruente)ion radunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40101834182232498322014-05-03T03:01:44.463-07:002014-05-03T03:01:44.463-07:00Since EA = ET = EB, FC = FT = FD and AB = CD.
A...Since EA = ET = EB, FC = FT = FD and AB = CD.<br />Also AD//BC so ABCD is an isosceles trapezoid. <br />Then EF = (AD + BC)/2. (see the solution of problem 1008).<br /> EF = AB = CD so AB + CD = AD + BC then ABCD is circumscribed quadrilateral . <br />Any isosceles trapezoid is a cyclic quadrilateral.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34295213211328144052014-04-26T04:45:58.329-07:002014-04-26T04:45:58.329-07:00Problem 1006 has been replaced. Thanks.Problem 1006 has been replaced. Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25423133476284975932014-04-25T18:14:50.549-07:002014-04-25T18:14:50.549-07:00This comment has been removed by the author.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1561668342074447702014-04-25T10:40:59.019-07:002014-04-25T10:40:59.019-07:00This comment has been removed by a blog administrator.Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22183603531831957942014-04-25T07:55:44.994-07:002014-04-25T07:55:44.994-07:00This comment has been removed by a blog administrator.Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88715253595030767122014-04-24T18:22:55.750-07:002014-04-24T18:22:55.750-07:00This comment has been removed by a blog administrator.Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17599528095406329222014-04-23T06:56:02.773-07:002014-04-23T06:56:02.773-07:00This comment has been removed by a blog administrator.Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com