tag:blogger.com,1999:blog-6933544261975483399.post1332074349865279047..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1389: Triangle, 40-120-20 degree, Angle, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-6933544261975483399.post-27196695141650246102024-01-31T14:29:20.361-08:002024-01-31T14:29:20.361-08:00thanks for you solution. How do you know that CD =...thanks for you solution. How do you know that CD = EF?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20465541570434909402020-04-05T10:59:03.835-07:002020-04-05T10:59:03.835-07:00More general form: A is twice of C.
The many solut...More general form: A is twice of C.<br />The many solution for 10-20.<br />https://artofproblemsolving.com/community/c6h487183p2729760Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62286424853546861232019-03-05T13:48:01.589-08:002019-03-05T13:48:01.589-08:00Hello !
Construct a triangle PCD outside the tri...Hello ! <br /><br />Construct a triangle PCD outside the triangle BDC , DPC=20° , PD=PC=BC.<br />PCB=60° and PBC is equilateral. So, PB=PD and BPD=40°. We conclude that DBP=BDP=70°. As we know, CBP=60°, so DBC=10°.<br />Finally, x=DBC+DCB=10°+20°=30°.Amiltonhttps://www.blogger.com/profile/08787118795734976763noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45301676989640999762019-02-21T23:27:50.167-08:002019-02-21T23:27:50.167-08:00Thank u for your kind words Dinesh. I thank God fo...Thank u for your kind words Dinesh. I thank God for whatever small talent I may have Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83463391136888322582019-02-21T12:39:37.620-08:002019-02-21T12:39:37.620-08:00Sumith Sir,
You are truly an expert at geometry p...Sumith Sir,<br /><br />You are truly an expert at geometry problems. You always provide very elegant solutions.Dinesh Nayak H.https://www.blogger.com/profile/11545154819645891319noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46504282539673987862018-10-03T05:01:11.947-07:002018-10-03T05:01:11.947-07:00Extend (AB to E so that triangle BCE be equilatera...Extend (AB to E so that triangle BCE be equilateral, take F on AC so that EF=EC, we get <EFC=<FCE=80, and <BEF=40, thus AF=EF and F=D, hence <AFB=<BEC/2=30.Stan FULGERhttps://www.facebook.com/stan.fulgernoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7717743176420158612018-09-27T15:49:45.742-07:002018-09-27T15:49:45.742-07:00Considering usual triangle notations,
let BC=AD=a ...Considering usual triangle notations,<br />let BC=AD=a and DC=x<br />Mark a point E on AC such that CE=a => AE=x ------------(1)<br />Observe that BEC is an 80-80-20 isosceles and ABE is a 40-40-100 isosceles with AE=BE=x -----------(2)<br />Now extend the internal angle bisector of m(ABC) to a point P such that BP=a and this forms an equilateral triangle BPC<br />From this construction it can be observed that C is the circumcenter of EBP <br />=> m(BEP) = (360-m(BCP))/2 = 360-60/2 = 150 degrees----------(3)<br />Another observation is that EB=DC=x, m(EBP)=m(DCB)=20 & BP=BC=a => EBP is congruent to DCB (SAS)<br />=>m(CDB)=m(BEP)=150<br />=>m(BDA)=x=30 Q.E.DSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14800790164831606462018-09-27T06:51:03.345-07:002018-09-27T06:51:03.345-07:00Luăm punctul E pe prelungirea lui BC astfel încât ...Luăm punctul E pe prelungirea lui BC astfel încât CE = CA; triunghiul AEC este isoscel (80-20-80).<br />Ducem BF paralel cu EA și atunci se arată ușor că EB=BF=FA=CD(triunghiul ABF isoscel 40-100-40).<br />Construim un triunghi isoscel CDP cu unghiul DPC = 20. <br />Deoarece CD = BF, implică triunghiul CBF = CDP și de aici CP = CB. De asemeni unghiul BCP = 60, ceea ce face triunghiul BPC echilateral(60-60-60) și PB =PC=PD. În triunghiul BDP isoscel BP = DP și <br /> unghiul BPD =40. De aici unghiul PDB = 70, adică unghiul BDC = 150 și deci BDA =30.<br />Florin Popa , Comanesti, Romania.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61676588357291532342018-09-27T02:16:35.907-07:002018-09-27T02:16:35.907-07:00Let point E between A and D such as AE=BE.So <...Let point E between A and D such as AE=BE.So <DBC=10.<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55357747278813321702018-09-26T19:48:22.063-07:002018-09-26T19:48:22.063-07:00Find E on AC such that AE = BE. Construct equilate...Find E on AC such that AE = BE. Construct equilateral Tr. BEF, F within Tr. BCE<br /><br />Since BC = AD = CE, AE = BE = EF = CD<br /><br />So Tr.s CEF & BCD are congruent SAS<br /><br />Hence < FCE = < CBD = 10 and so<br />x = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com