tag:blogger.com,1999:blog-6933544261975483399.post7925617265656894714..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #12Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-74091720150164403992024-03-26T00:53:21.678-07:002024-03-26T00:53:21.678-07:00Hello Sumith.
In your proof, the sentence "DS...Hello Sumith.<br />In your proof, the sentence "DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide" seems to infer that SC, CT and TS should all be positive or equal to 0.<br />How can we sure about that ?<br />Best regards.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87544674751176135402024-03-22T09:41:03.357-07:002024-03-22T09:41:03.357-07:00Hi Peter!
You mean "AD cut BE at L ( see pict...Hi Peter!<br />You mean "AD cut BE at L ( see picture)" ?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32205659669360023292018-03-29T02:55:38.180-07:002018-03-29T02:55:38.180-07:00Link doesnt work
Link doesnt work<br />Anonymoushttps://www.blogger.com/profile/03704233848451554554noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43661388451072297202015-11-06T01:51:02.985-08:002015-11-06T01:51:02.985-08:00Let AD, BE meet at M. MF is obviously perpendicula...Let AD, BE meet at M. MF is obviously perpendicular to AB<br /><br />Let DC meet MF at S and EC meet MF at T<br /><br />< SDM = < SMD = 90 - A so DS = SM ....(1)<br />Similarly MT = TE .....(2)<br />DC = CE......(3)<br /><br />(1) +(2) -(3) yields DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide<br /><br />So C must lie on MF which completes our proof<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36170507564482065792011-08-06T23:43:59.477-07:002011-08-06T23:43:59.477-07:00Peter Tran's solution gives me this idea:
To c...Peter Tran's solution gives me this idea:<br />To construct the tangents we draw the circle with diameter CO<br />This cercle intersects the given semi circle at D and E<br />he passes trough O midpoint of AB and the altitude feet E,D, he is the nine point circle of triangle ABL<br />let be M the second intersection with AB<br />M is an altitude foot<br />CM perpendicular to AB (CO diameter)<br />hence CM is an altitude and passes trough orthocenter F and M=G<br />FC perpendicular to ABAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76534477759401285152011-08-06T22:57:54.097-07:002011-08-06T22:57:54.097-07:00Peter Tran's solution suggests the following (...Peter Tran's solution suggests the following (perhaps easier) problem:<br /><br />"Let D, E be any two points on the semicircle on AB as diameter. Let AE, BD intersect at F; AD, BE intersect at L. Prove LF is perpendicular to AB."<br /> <br />PravinPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85238057001830588622011-08-04T19:18:02.815-07:002011-08-04T19:18:02.815-07:00Nice solution!Nice solution!Slicehttp://www.fmat.clnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39593518922979372992011-08-03T21:26:01.845-07:002011-08-03T21:26:01.845-07:00http://img193.imageshack.us/img193/4671/problem652...http://img193.imageshack.us/img193/4671/problem652.png<br /><br />Connect OC , DE<br />AD cut BF at L ( see picture)<br />Quadrilateral LDFE is cyclic with LF as diameter ( Angles D=E = 90)<br /> Note that angle(DAE)= angle (CDE)<br />So angle (DLE)= angle (DCO) ( angles complement to congruence angles)<br />OC is perpendicular bisector of DE . On OC only center of circle LDFE have property (DLE)= ½ (DCE)<br />So C is the center of circle LDFE and diameter LF pass through center C<br />In Triangle ABL , F is the orthocenter so line LCF will perpendicular to AB<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com