tag:blogger.com,1999:blog-6933544261975483399.post1398229256528097622..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1244: Circle, Radius, Perpendicular, Chord, Secant, Measurement. Mind MapAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-78306962225478592382021-01-21T09:37:31.929-08:002021-01-21T09:37:31.929-08:00First you will need to prove the collinearity of F...First you will need to prove the collinearity of F.O.DAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56853275178648421642021-01-20T10:19:46.513-08:002021-01-20T10:19:46.513-08:00Join points F.O.D then WE CAN eaisly solve for r.
...Join points F.O.D then WE CAN eaisly solve for r.<br />Note that FO=OD=OC=r so Triangle FCD is a right triangle <br />By similar angle WE have traingle FCD is similar to ECA so<br />4/3+x=3/10 then X=31/3Anonymoushttps://www.blogger.com/profile/04652001034659981242noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68406315148444615632016-08-14T10:47:28.902-07:002016-08-14T10:47:28.902-07:002nd solution
< CFD = < CGD = < EAD as b...2nd solution <br /><br />< CFD = < CGD = < EAD as before <br /><br />Hence AFED is cyclic and the result is easily calculated <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15607338849700906342016-08-14T05:47:03.362-07:002016-08-14T05:47:03.362-07:00Let AB cut the circle at G.
In isoceles Tr. BCG,...Let AB cut the circle at G. <br /><br />In isoceles Tr. BCG, <br /><br />CG^2 - CE^2 = BE.GE = CE.EF = 24<br />So CG^2 = 24 + 16 = 40<br /><br />Now < CGB = CBG = < GDA <br /><br />Hence < CGD = < DGB - < CGB = < DGB - < GDA = < GAD<br /><br />Hence CG^2 = CD.CA<br /><br />So 40 = 3(3+x) from whence x = 31/3<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23438815823883738632016-08-13T22:26:55.586-07:002016-08-13T22:26:55.586-07:00https://goo.gl/photos/FGTGW9iSkgxRKfx16
Connect B...https://goo.gl/photos/FGTGW9iSkgxRKfx16<br /><br />Connect BF, BC and BD<br />Note that C is the midpoint of arc BG<br />Triangle BEC similar to FBC ( case AA)<br />So CB^2=CE.CF<br />Triangle DCB similar to BCA ( case AA)<br />So CB^2=CD.CA<br />And CE.CF=CD.CA or 4x 10= 3(x+3)<br />So x= 31/3 <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com