tag:blogger.com,1999:blog-6933544261975483399.post6372616803218340273..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1254: Circle, Arc, Chord, Midpoint, Cyclic Quadrilateral, Concyclic PointAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-37208462548923336622016-08-31T21:43:59.903-07:002016-08-31T21:43:59.903-07:00Problem 1254
Is <GBA=<GAB=<DAB (AF=FB, F...Problem 1254<br />Is <GBA=<GAB=<DAB (AF=FB, FG=perpendicular bisector of AB),<BAD=arcBD=arcDE.<br />Now <CHG+<CBG=(arcCB+arcBG+arcAE)/2+arcCA/2+<BAD=<br />(arcCB+arcBG+arcAE+arcCA+arcDE)/2=360/2=180.<br />Therefore CBGH is cyclic. 3<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3118575012522398452016-08-31T19:18:58.613-07:002016-08-31T19:18:58.613-07:00Join AE and form the triangle AHE.
Let Angle BAD ...Join AE and form the triangle AHE. <br />Let Angle BAD = A => Angle DAE = Angle HAE = A (since D is midpoint of B & E)<br />Let Angle ABC = B => Angle CEA = Angle HEA = B => Angle AHE = Angle CHG = 180-(A+B)<br />and since C is midpoint of AB => Angle CAB = B and triangle ABC is isosceles <br />=> CF perpendicular to AB and ABG is isosceles => Angle ABG = A<br />Therefore Angle CBG = A+B & Angle CHG = 180-(A+B) => BCGH are concyclicAnonymousnoreply@blogger.com