tag:blogger.com,1999:blog-6933544261975483399.post3927084750475906841..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 536: Intersecting Circles, Chord, Perpendicular. Level: High School, SAT Prep, College GeometryAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-62003359913740117412010-11-03T16:07:36.744-07:002010-11-03T16:07:36.744-07:00Let E is the point of intersection of AC to circle...Let E is the point of intersection of AC to circle O.<br />Connect OO’ and note that OO’ is the perpendicular bisector of AB<br /><br />1. We have Angle(BEC)=angle (O’OA) <br />And angle(ACB)=2 *angle(O’OA)=angle(BEC)+angle(CBE) <br />So angle(CBE)=angle(O’OA)=angle(CBE) <br />Triangle BCE isosceles and CB=CE<br /><br />2. Since AE is a chord of circle O so D is the midpoint of AE<br />We have Area(ADO)=Area(DOC)+Area(COE)<br />All above triangles have the same height so AD=DC+CE <br />Or AD=BC+CB<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48252250412441452392010-11-03T14:09:23.915-07:002010-11-03T14:09:23.915-07:00extend AC to G on circle O, => BCG isoceles
=&g...extend AC to G on circle O, => BCG isoceles<br />=> AD = DC + CG = BC + BCc .t . e. onoreply@blogger.com