tag:blogger.com,1999:blog-6933544261975483399.post6282027294814141207..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 688: Triangle, Angles, 10, 20, 30, 40, 60 Degrees, Measure, Mind Map, PolyaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-7519895235951571992018-09-29T07:02:05.871-07:002018-09-29T07:02:05.871-07:00We are asked to PROVE it using Euclidean principle...We are asked to PROVE it using Euclidean principlesSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17811193619823750102018-02-07T12:25:07.799-08:002018-02-07T12:25:07.799-08:00When I constructed the original drawing on Geogebr...When I constructed the original drawing on Geogebra without any extra parts, the measurement of the angle was 20 degrees. Geek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72618861626170048622017-11-18T06:09:00.254-08:002017-11-18T06:09:00.254-08:00You draw a parallel to AC thro' B and then u d...You draw a parallel to AC thro' B and then u draw a circle (D,DB) to intersect the parallel at say P.<br /><br />Your proof then assumes that D,E,P are collinear -this is by no means evident or axiomatic. It must be PROVED.<br /><br />Antonio -I would like u to pls comment too<br /><br />Rgds<br /><br />Sumith PeirisSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12047204639908720152017-11-06T11:02:58.941-08:002017-11-06T11:02:58.941-08:00Here's my solution
https://www.youtube.com/wa...Here's my solution<br /><br />https://www.youtube.com/watch?v=6DCTArdbSpEGeek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76976871532459792712016-06-27T03:53:16.697-07:002016-06-27T03:53:16.697-07:00Problem 688
Draw the equilateral triangles DEK, t...Problem 688<br />Draw the equilateral triangles DEK, then <EBD=<EKD=60.So EBKD is cyclic.Then <DEK=<DBK=60=<DBC.Therefore B,K,C are collinear. Is <EKD=60=2.30=2.<ECD,<br />Then the point K is circumcenter of triangle DEC .Then x=<EDB=<EKB=10+10=20.<br />MANOLOUDIS APOSTOLIS FROM GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86202275734708472172015-11-10T10:21:24.274-08:002015-11-10T10:21:24.274-08:00Or simply EF = FC = DC and CDEF is concyclic so x+...Or simply EF = FC = DC and CDEF is concyclic so x+30 = 10+40 hence x= 20Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5411391132035305142015-08-25T01:48:52.357-07:002015-08-25T01:48:52.357-07:00Find F on AB extended such that BC bisects < AC...Find F on AB extended such that BC bisects < ACF. Then Tr. EFC is congruent with Tr. DFC and so EC = EF from which we can prove that Tr.s AFD and ACE are congruent. Hence AE = AD and so x = 20<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37316621332220951832012-10-19T16:57:41.692-07:002012-10-19T16:57:41.692-07:00Reflection in the line CE. Let D→D'.
Then ΔD...Reflection in the line CE. Let D→D'. <br /><br />Then ΔDCD' is equilateral. <br /><br />∵ ∠DBC = ∠DD'C = 60°<br />∴ B, D, C, D' are concyclic. <br /><br />∵ ∠EBD = ∠DCD' = 60°<br />∴ E, B, D' are collinear. <br /><br />In ΔACD', ∠AD'C = 100°<br />∴ ∠ED'D = 40°<br /><br />∵ ED = ED'<br />∴ ∠EDD' = 40°<br /><br />∵ ∠BDD' = ∠BDD' = 20°<br />∴ x = ∠EBD = 20°Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68376948544532320772012-10-19T08:20:14.887-07:002012-10-19T08:20:14.887-07:00Reflection in the line BC. Let D→D'.
∵ ∠DBC ...Reflection in the line BC. Let D→D'. <br /><br />∵ ∠DBC = ∠D'BC = 60°<br />∴ E, B, D' are collinear. <br /><br />∵ ∠BDC = ∠BD'C = 80°<br />∴ ∠D'CA = 80°, ∠D'CB = 40°<br /><br />Join DD'. Then BD = BD', <br />∴ ∠BDD' = 30°, ∠D'DC = 50°<br /><br />∵ ∠D'DC = ∠D'EC = 50°<br />∴ C, D, E, D' are concyclic. <br /><br />∴ ∠D'DE = ∠D'CE = 50°<br />∴ x = ∠BDE = ∠D'DE − ∠BDD' = 20°Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44430654434002393082011-12-22T02:03:51.327-08:002011-12-22T02:03:51.327-08:00Did you both used AD=AE which isn't given or h...Did you both used AD=AE which isn't given or how can someone conclude EF=CD?SCPnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79662099318597877552011-11-18T15:43:05.003-08:002011-11-18T15:43:05.003-08:00Extend AB till F so that AF=AC. This way DCFE is a...Extend AB till F so that AF=AC. This way DCFE is an isosceles trapezoid(CD=CF=EF and <DCF=<CFE=80). <AED =80=<EBD+<BDE. So <BDE=20.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26763345202521585382011-11-16T22:26:37.348-08:002011-11-16T22:26:37.348-08:00http://img403.imageshack.us/img403/6706/sfvdf.png
...http://img403.imageshack.us/img403/6706/sfvdf.png<br /><br />Extend AB from B till F so that < EFD = 30°. Since < CEF = 50°, then < CDF = 50° and DF ⊥ BC, whereat < BDF = 30°, and finally from the cyclic quadrilateral CDEF we have x + 30° = 50° and x = 20°.Alejandrohttp://www.fmat.clnoreply@blogger.com