tag:blogger.com,1999:blog-6933544261975483399.post301196192531950087..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 421: Right Triangle, Cevian, Angles, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-88865250825020026532020-05-02T20:12:54.635-07:002020-05-02T20:12:54.635-07:00It is special case of more general problem: (30-2t...It is special case of more general problem: (30-2t, 120-3t, 60-2t, 30) where t = 10.<br /><br />The problem is related with following cevian problem.<br />It can be generated by taking P on EC where AEDP concyclic.<br />https://output.jsbin.com/fofecum#100,10,10,30Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48203847973513669732016-08-16T01:01:31.351-07:002016-08-16T01:01:31.351-07:00Problem 421
The circle (D, DC) intersects the AC,...Problem 421<br />The circle (D, DC) intersects the AC, CE in K, L respectively then <LDK=2<LCK=2.30=60.So<br />Triangle KDL is equilateral .Is <DLC=<DCL then <ELK=<EAK=50 so AKEL is cyclic. But <KLD=60=<br />2.30=2<KAD therefore the K is circumcenter the triangle AKD so LA=LK=LD=KD=DC and <br /><LAK=<LKA=50+30=80 or <LKE=<LAE=80-50=30.Then triangle LKE=triangle DKE so LE=ED and<br /><EDL=10,so <EDA=<BDA-<BDE=70-(20+10)=40.<br />Αnother solution the circle (E,EA) similar.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41194620173303066452015-12-29T04:36:02.978-08:002015-12-29T04:36:02.978-08:00Pure Geometry proof
AD, CE meet at F. Since AF = ...Pure Geometry proof<br /><br />AD, CE meet at F. Since AF = CF let MF the perpendicular bisector of AC meet BC at G and AB extended at H<br /><br />Now BC is the perpendicular bisector of EH and since it therefore bisects < ECH and DF bisects < CFH (= 120) D is the incentre of Tr. FHC<br /><br />Hence <DHG = 20, < BDH = BDE = 30 and so x must be < 40 since < ADB = 70<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12504705366385715952014-08-18T09:57:07.768-07:002014-08-18T09:57:07.768-07:00Let circumcircle of EDC meet DA at F.
Triangle FEC...Let circumcircle of EDC meet DA at F.<br />Triangle FEC forms a trigonometric cevian model.<br />From angle C in clockwise, we get 60-x,10,10,100,30,x-30.<br />This is the classical 10-10-10-20-100-30 problem which makes x-30=10 => x=40.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43142741258076508082010-03-20T08:51:51.322-07:002010-03-20T08:51:51.322-07:00To Anonymous (March 19, 2010 10:55 PM):
The answer...To Anonymous (March 19, 2010 10:55 PM):<br />The answer is not correct.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83416571291682529592010-03-19T22:55:27.140-07:002010-03-19T22:55:27.140-07:00Set A be Centre,
CED=30/2=15
CED+x=60
x=45Set A be Centre,<br />CED=30/2=15<br />CED+x=60<br />x=45Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42198910476584090372010-01-26T13:58:17.557-08:002010-01-26T13:58:17.557-08:00Why AB=BCtan40. Anybody can help.Why AB=BCtan40. Anybody can help.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4557471635326945262010-01-26T06:02:00.669-08:002010-01-26T06:02:00.669-08:00To Anonymous:
The answer is not correct.To Anonymous:<br />The answer is not correct.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-66388519521181217272010-01-26T05:39:09.537-08:002010-01-26T05:39:09.537-08:00angle B + angle A + angle C =180
90 + 50 + angle E...angle B + angle A + angle C =180<br />90 + 50 + angle ECA + 10 = 180<br />hence, angle ECA = 30<br />now,<br />angle A + angle ECA + angle AEC = 180<br />50 + 30 + angle AEC = 180<br />angle AEC = 100<br />now,<br />angle EAD + angle AED + angle ADE = 180<br />20+100+angle ADE(or x)= 180<br />hence,<br />x = 60Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20382014038283298712010-01-23T09:32:27.492-08:002010-01-23T09:32:27.492-08:00AB=BCtan40
BD=ABtan20
EB=BCtan10
tan(EDB)=tan10/(t...AB=BCtan40<br />BD=ABtan20<br />EB=BCtan10<br />tan(EDB)=tan10/(tan20.tan40)=tan30<br />x=70-30=40<br />.-.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33644697396191603122010-01-21T20:41:13.960-08:002010-01-21T20:41:13.960-08:00http://geometri-problemleri.blogspot.com/2010/01/p...http://geometri-problemleri.blogspot.com/2010/01/problem-63-ve-cozumu.htmlAnonymousnoreply@blogger.com