tag:blogger.com,1999:blog-6933544261975483399.post3187234818329822224..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 574: Circle, Chord, Secant, Angle Bisector, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-29816185988944590072015-07-14T03:21:23.297-07:002015-07-14T03:21:23.297-07:00Let G cut BC at X and AD at Y. Then < A = <...Let G cut BC at X and AD at Y. Then < A = < C. Hence Tr. FXY is isoceles 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-27380905642105859332011-01-27T04:23:32.294-08:002011-01-27T04:23:32.294-08:00Problem #574, To c.t.e.o: Thanks.Problem #574, To c.t.e.o: Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69342279055187549612011-01-27T04:05:03.110-08:002011-01-27T04:05:03.110-08:00To Antonio,Thanks. H, E, P are never collinear
HEP...To Antonio,Thanks. H, E, P are never collinear<br />HEP is the same bisector for vertcales angles<br />We have to prove bisector of P // to bisector of CBE<br />AC, DB meet at P1, bisector of CEB meet DP1 at P<br />bisector of P1 meet CD at E1<br />In ▲P1CE1 1/2 P1 + P1CE1 + CE1P1 = 180<br />In E1EPP1 180- CE1P1 + E1EP + E1EP + PBE + 1/2P1=360<br />from both => 2CE1P1 = 2E1EP => CE1P1 = E1EP<br />as corresponding angle => P1E1//PEc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15167027895886163752011-01-26T14:10:04.215-08:002011-01-26T14:10:04.215-08:00Problem #574. To c.t.e.o: Thanks for your comment....Problem #574. To c.t.e.o: Thanks for your comment. What is the reason for your statement HEP is the bisector of CEB (why are points H, E, and P collinear?). Thanks.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14367402765625073082011-01-26T08:12:42.452-08:002011-01-26T08:12:42.452-08:00To Antonio
HEP bisector of CEB, AEC = DEB
AEC + 1...To Antonio<br /><br />HEP bisector of CEB, AEC = DEB<br />AEC + 1/2 CEB = DEB + 1/2 CEBc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41664178908150430042011-01-26T05:34:14.547-08:002011-01-26T05:34:14.547-08:00Problem #574. To c.t.e.o: What is the reason for y...Problem #574. To c.t.e.o: What is the reason for your statement angle AEP = angle DEP?Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17673537181025498902011-01-26T01:42:50.756-08:002011-01-26T01:42:50.756-08:00name P intersection of AC, DB extended
▲AEP ~ ▲DEP...name P intersection of AC, DB extended<br />▲AEP ~ ▲DEP ( CAB = CDB, AEP = DEP )<br />=> APH = DPH => PH bisector from 572<br />PH perpendicular to FHc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89199400949075122462011-01-26T01:07:47.891-08:002011-01-26T01:07:47.891-08:00http://img137.imageshack.us/img137/1631/problem574...http://img137.imageshack.us/img137/1631/problem574.png<br /><br />EH cut AD at K and BC at L ( see attached picture)<br />Note that Triangle KED similar to tri. LEB (case AA)<br />So angle(LKF)=angle(KLF) and triangle KLF is isosceles<br />And angle bisector FH of isosceles triangle KLF perpendicular to the base KL at H<br /><br />Peter Trancltran34https://www.blogger.com/profile/08094245599507131157noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50889416132130974452011-01-25T18:13:56.340-08:002011-01-25T18:13:56.340-08:00Let FM intersection CD be G
Denote the "acute...Let FM intersection CD be G<br />Denote the "acute" angles at <br />A, B, C, D, E, F, G<br />by the corresponding small letters <br />a, b, c, d, e, f, g.<br />b = d = f + c <br />f = b – c <br />d = g + (f/2) , <br />g = d – (f/2) = b – [(b-c)/2] = (b + c)/2<br />e + b + c = Pi,<br />(e/2) = (Pi/2) – [(b+c)/2] = (Pi/2) – g <br />(e/2) + g = Pi/2 in triangle EHG<br />So angle at H = 90 degreesPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com