tag:blogger.com,1999:blog-6933544261975483399.post9134219571625065431..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 745: Circle, 90 Degrees, Chord, Congruence, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-66833098105955109292016-07-01T05:25:16.675-07:002016-07-01T05:25:16.675-07:00Problem 745
From the theorem of broken chord (A...Problem 745<br />From the theorem of broken chord (Archimedes) is arc AB=arc BD then AE=EC+CD.<br />So 5=2+x .Therefore x=3<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65448354811941226322015-09-06T13:17:42.938-07:002015-09-06T13:17:42.938-07:00Still another method much simpler than both previo...Still another method much simpler than both previous proofs...,<br /><br />Find G on AE such that GE = 2. Then prove that Tr.s AGB & BDC are congruent SAA and the result follows. <br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58142060466603331312015-09-06T13:06:21.579-07:002015-09-06T13:06:21.579-07:00Another proof without using Pythagoras...
Extend ...Another proof without using Pythagoras...<br /><br />Extend EC to F such that CF = 3. So Tr. ABF is isoceles and hence B is the centre of circle ADF. Hence < AFD = 1/2 of < ABD = 45 and so x = CF = 3<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32773016824658148822015-09-06T12:56:10.078-07:002015-09-06T12:56:10.078-07:00AD is a diameter so < BDA = 45 = < BCA. Henc...AD is a diameter so < BDA = 45 = < BCA. Hence BE = EC = 2 and so applying Pythagoras to Tr. ABE, AB^2 = 29 from which we have AD^2 = 58. <br />So x^2 = 58 - 7^2 = 9 and thus x = 3<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62510254727731767022012-07-04T14:37:42.893-07:002012-07-04T14:37:42.893-07:00http://img18.imageshack.us/img18/6838/p745resuelto...http://img18.imageshack.us/img18/6838/p745resuelto.png<br /><br />(look at the picture)<br /><br />Note that ABCD is cyclic, then 7*BD=BD*x+2sqrt(2)*BD*sqrt(2)<br /><=>x=3<br /><br />GreetingsEder Contreras Ordeneshttps://www.blogger.com/profile/08915905860162020945noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76149055723529444652012-05-02T04:33:25.639-07:002012-05-02T04:33:25.639-07:00B is the midpoint of arc AD and BE is perpendicula...B is the midpoint of arc AD and BE is perpendicular to chord AC. Hence by the Broken Chord Theorem, 5 = 2 + x or x = 3<br />If you wish to prove the theorem all you need to do is extend AC to F such that CF =CD=x. Now /_ABD = /_ACD = 2*/_AFD and B being the midpoint of arc AD lies on the perpendicular bisector of AD. Therefore B is the circumcentre of triangle ADF and BE is perpendicular to AF so AE = EF or 5 = 2 + x or x = 3<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8467387915838775062012-04-30T08:49:07.778-07:002012-04-30T08:49:07.778-07:00http://img715.imageshack.us/img715/7306/problem745...http://img715.imageshack.us/img715/7306/problem745.png<br /><br />Since ∠ACD=90 => AD is a diameter<br />Since AB=BD => BO perpendicular to AD<br />Let BE cut AD at F and r = radius of the circle<br />Since EF//CD => AE/AC=AF/AD => AF=10/7 .r<br />Note that ∠EAF=∠OBF= alpha<br />In ∆ BOF , tan(alpha)=OF/OB=3/7<br />In ∆CAD, tan(alpha)=x/7=3/7<br />So x=3Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77093575458542369292012-04-30T05:50:35.800-07:002012-04-30T05:50:35.800-07:00The solution is uploaded to the following link:
h...The solution is uploaded to the following link:<br /><br />https://docs.google.com/open?id=0B6XXCq92fLJJM0w2aWEyS3lMNEkAnonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36912807120714324142012-04-30T05:39:15.996-07:002012-04-30T05:39:15.996-07:00The slution is uploaded to the following link:
ht...The slution is uploaded to the following link:<br /><br />https://docs.google.com/open?id=0B6XXCq92fLJJNkJFWExzWE1NSW8Anonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.com