tag:blogger.com,1999:blog-6933544261975483399.post204623788446245674..comments2022-11-29T01:21:52.908-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1071: Arbelos, Semicircles, Diameter, Circle, Perpendicular, Common Tangent, Parallel, Collinear Points, Midpoint, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-40697125252964201172014-12-30T04:44:50.239-08:002014-12-30T04:44:50.239-08:00Let BD and A₁C₂ intersect at P.
Join A₁B and BC₂...Let BD and A₁C₂ intersect at P. <br /><br />Join A₁B and BC₂. Then A₁BC₂D is a rectangle. <br />Thus PA₁=PB=PC₂=PD, and so A₁C₂ is the common tangent of O₁ and O₂. <br /><br />***<br />Clearly, A₁B//A₃O₂//DC, and BO₂=O₂C, so A₁A₃=A₃D. <br />Similarly, C₂C₄=C₄D. Therefore, A₁C₂//A₃C₄ and A₁C₂=2×A₃C₄. <br /><br />Since A₁BC₂D is rectangle, thus A₁C₂=BD=2×A₃C₄. <br />Also since P is mid-point of A₁C₂, so M is mid-point of A₃C₄. <br /><br />***<br />Since A₁O₁//A₃B//DO₄, and also A₁C₂//A₃C₄, <br />so A₃C₄ is tangent to circle O₂. <br /><br />Similarly A₃C₄ is tangent to circle O₄. <br />And hence A₃C₄ is the common tangent of O₂ and O₄. <br /><br />Since BD passes through M, and M is mid-point of A₃C₄, <br />thus M lies on the radical axis of O₂ and O₄, which is BD. <br />Therefore BD passes through E, which also lies on the radical axis of O₂ and O₄. <br />Hence, B,E,D are collinear. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com