tag:blogger.com,1999:blog-6933544261975483399.post5550004161348651591..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 358. Isosceles triangle 80-80-20, Circle, Angles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-39586309328869684482021-01-21T09:28:46.708-08:002021-01-21T09:28:46.708-08:00Rotate the triangle ACD (60-80-40) clockwise over ...Rotate the triangle ACD (60-80-40) clockwise over C by 20 degrees such that AC overlaps EC. Denote the new position of D as P (E is the new position of A). Let CP intersect the pink circle at Q. <br />We can see that ECF is an equilateral triangle and AEC is an 80-80-20 triangle. <br />Since m(ECP)=m(ACD)=80 and m(ECF)=60=> m(DCP)=20 and as DC=CP per construction => DCP is a 80-20-80 isosceles triangle<br />As CF=CQ=Radius of the circle => DF=PQ and FQ//DP. On angle chasing we can see that FPQ is a 40-100-40 isosceles and FQ=PQ<br />Since FCQ is an 80-20-80 isosceles and FC=AC=CQ=EC => FCQ is congruent to ACE<br />=> AE=FQ=PQ=DFSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83133891991505128142016-08-12T14:25:33.981-07:002016-08-12T14:25:33.981-07:00Problem 358
Triangle ECF is equilateral. The EF i...Problem 358<br />Triangle ECF is equilateral. The EF intersects AC in P (P is left of A ).Then <PEA=<EPA=40 so PA=AE.But triangle ACD= triangle FCP (AC=CF, <CPF=40=ADC,<PCF=<ACD=80). Therefore<br />CD=PC or CF+FD=CA+AP so DF=AP=AE.<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-13670513749544142582015-10-05T02:41:45.980-07:002015-10-05T02:41:45.980-07:00Let AD cut the circle at P and EF at Q.
Easily T...Let AD cut the circle at P and EF at Q. <br /><br />Easily Tr. AEC is 80-20-80 and so Tr. CEF is equilateral. Hence < DAF = 1/2 *60 - 20 = 10 and similarly < EFA = 1/2 *< ECA = 10. So AEPF is an isoceles trapezoid with AE = PF<br /><br />Now since < ACF = 80, < PFD = 140 and since AD = BD, < PDF = DPF = 40. Hence PF = DF and so AE = DF<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13437703668274365462010-07-02T02:33:47.390-07:002010-07-02T02:33:47.390-07:00http://ahmetelmas-geo-geo-antonio.blogspot.com/http://ahmetelmas-geo-geo-antonio.blogspot.com/Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47274958752993423702009-11-08T06:36:18.089-08:002009-11-08T06:36:18.089-08:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.com