tag:blogger.com,1999:blog-6933544261975483399.post3537538116917440477..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1241: Quadrilateral, Four Exterior Angle Bisectors, Cyclic Quadrilateral, CircleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-79844169926654428592016-08-03T17:03:24.563-07:002016-08-03T17:03:24.563-07:00Correction - this will be only if ABCD is cyclicCorrection - this will be only if ABCD is cyclicSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43552558191909189012016-08-02T07:43:17.160-07:002016-08-02T07:43:17.160-07:00It can be proved that the diagonals of this cyclic...It can be proved that the diagonals of this cyclic quadrilateral are perpendicular to each other. Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49405273252165261872016-08-02T00:46:32.338-07:002016-08-02T00:46:32.338-07:00Problem 1241
Set <A1AB=<D1AD=x,<A1BA=&l...Problem 1241<br />Set <A1AB=<D1AD=x,<A1BA=<B1BC=y,<B1CB=<DCC1=z and <CDC1=<ADD1=φ.Then <br />2x+2y+2z+2φ=180.But <AA1B+<DC1C=180-(x+y)+180-(z+φ)=180.Therefore A1, B1, C1 <br />and D1 are concyclic.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36342173737668913682016-08-01T22:59:32.689-07:002016-08-01T22:59:32.689-07:00Let ext∠A=2a, similarly for B, C, D.
a+b+∠A₁=180...Let ext∠A=2a, similarly for B, C, D. <br /><br />a+b+∠A₁=180°<br />c+d+∠C₁=180°<br />a+b+c+d+∠A₁+∠C₁=360°<br /><br />But a+b+c+d=180°<br />Thus ∠A₁+∠C₁=180°<br />Hence A₁B₁C₁D₁ is concyclic. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5267736234666924712016-08-01T21:57:57.751-07:002016-08-01T21:57:57.751-07:00< A1AB = 90 - A/2 and <A1BA = 90 - B/2
Hence...< A1AB = 90 - A/2 and <A1BA = 90 - B/2<br />Hence <A1 = A/2+B/2….(1)<br /><br />Similarly we can show that <br /> <C = C/2+D/2….(2)<br /><br />So <A1+<C1 = (A/2+B/2)+(C/2+D/2) = (A+B+C+D)/2 = 360/2 =180<br /><br />Hence A1B1C1D1 is a cyclic quadrilateral.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com