tag:blogger.com,1999:blog-6933544261975483399.post6905701037879747788..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 713: Tangent Circles, Diameter, Chord, Tangent, Incenter, Incircle, TriangleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-33462372811596305202012-01-04T23:13:03.649-08:002012-01-04T23:13:03.649-08:00Referring to Problem 712, recall that
x = (90° + 2...Referring to Problem 712, recall that<br />x = (90° + 25°)/2,<br />more generally, if we denote ∠CAB by θ we can <br />prove similarly that x = (90° + θ)/2<br />Applying it to the the present Problem 713,<br />∠CIB = (90° + θ)/2 <br />Follows that ∠ICE = (90° - θ)/2 = ∠ACE /2<br />So CI bisects ∠ACE.<br />Already AB bisects ∠CAE<br />Hence I is the incentre of ∆ACE.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9825971497003738462012-01-04T20:57:28.497-08:002012-01-04T20:57:28.497-08:00http://img12.imageshack.us/img12/9500/problem713.p...http://img12.imageshack.us/img12/9500/problem713.png<br />Connect O’D, BC and IC<br />We have O’D perpen. AC and BC perpen. To AC<br />Let H is the projection of O’ over BC (see picture)<br />DCHO’ is a rectangle and tri. IDO’ congruence to tri. HO’B ----(case ASA)<br />So O’H=DC=DI >>> Tri. IDC is isosceles<br />Angle (DIC)=(ICE)=(DCI)----- (alternate angle)<br />So CI is an angle bisector of tri. ACE and I will be incenter of tri. ACEPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com