tag:blogger.com,1999:blog-6933544261975483399.post5776846115139250871..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1236: Intersecting Circles, Secant, Concyclic PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-49727218197493525642016-09-17T10:30:20.256-07:002016-09-17T10:30:20.256-07:00https://goo.gl/photos/1g61g7m5f8PYXRTeA
Connect P...https://goo.gl/photos/1g61g7m5f8PYXRTeA<br /><br />Connect PQ, EB, BD, PB and QB ( see sketch)<br />Note that u= ∠ (BEQ)= ∠ (APQ)= ∠ (BPQ)<br />So E, P, B and Q are cocyclic.<br />Similarly P, B, Q and D are cocyclic<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12007643322814708772016-09-16T04:06:36.516-07:002016-09-16T04:06:36.516-07:00The solution that I propose is the following:
<...The solution that I propose is the following:<br /><PEA=<ADQ=<EAQ=<DAQ=x ; <EPA=AQD=y= 180-2x<br /><PAB=<PBA=α ; <APB=K=180-2α<br /><QAB=<ABQ=β ;<AQB=R=180-2β ; <ABQ+<ABP = α + β<br />To be concyclic the points BPED and Q need that:<br /><R+<K+<y=180 or α + β +x= 180 ; <br /><R+<K+<y=180-2β+180-2α+180-2x=α + β +x=180<br />Therefore point BPED and Q are concyclic <br />Reply<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58912311729099493762016-07-22T13:43:35.762-07:002016-07-22T13:43:35.762-07:00Problem 1236
Is <EPA=2.<EBA (PE=PA), <A...Problem 1236<br />Is <EPA=2.<EBA (PE=PA), <AQD=2.<ABD (AQ=QD).But <EAP=<DAQ=x. Then <EPD=<EPA=<br />180-2x=2(90-x)=<AQD=<EQD.But <EBD=<EBA+<ABD=(< EPA)/2+(<AQD)/2=90-x+90-x=180-2x.Therefore points E,P,B,Q and D are concyclic.<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-36280233281617132662016-07-16T20:28:16.975-07:002016-07-16T20:28:16.975-07:00Isosceles triangles PEA and QDA. Since <PBQ+<...Isosceles triangles PEA and QDA. Since <PBQ+<E=<PAQ+<EAP=180, E,P,B,Q concyclic. Similary, D,Q,B,P concyclic. Therefore, E,P,B,Q,D concyclic.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37662341095398356162016-07-15T10:39:55.546-07:002016-07-15T10:39:55.546-07:00< PAQ = < PBQ = 135. How did u arrive at thi...< PAQ = < PBQ = 135. How did u arrive at this conclusion?Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57124945216275073052016-07-14T14:15:32.645-07:002016-07-14T14:15:32.645-07:00<PAQ = <PBQ = 135 which are supplimentary. H... <PAQ = <PBQ = 135 which are supplimentary. Hence B also lies on the same circle. For some reasons, this did not get printed in my first comment. Faced this trouble couple of times and I stopped posting. ( I have posted <PEA = <PEQ = 45 in the beginning, but the preview is truncating this as well)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83323740197352547422016-07-14T09:55:02.881-07:002016-07-14T09:55:02.881-07:00connect PB, BQ, <PBQ = <PAQ,
<PAQ + <P...connect PB, BQ, <PBQ = <PAQ,<br /><PAQ + <PAE = 180 = <PAQ + <PEA =< PBQ + <PEA =180,<br />SO b ALSO LIES ON THE SAME CIRCLEAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-8471124042483145712016-07-13T21:21:41.054-07:002016-07-13T21:21:41.054-07:00"B also lies on the same circle"
Could ..."B also lies on the same circle"<br /><br />Could u pls explain how u arrived at this conclusion? Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76996962071657778932016-07-13T18:55:31.711-07:002016-07-13T18:55:31.711-07:00Isosceles triangles PEA and QDA are similar => ...Isosceles triangles PEA and QDA are similar => EA/PA = DA/QA => EA.AQ = PA.AD <br />Hence E,P,Q and D are concyclic. <br /><br /> B also lies on the same circle Anonymousnoreply@blogger.com