tag:blogger.com,1999:blog-6933544261975483399.post9028346597064318445..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 705: Intersecting Circles, Diameter, Angles, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-58507411980319758562012-02-19T07:00:58.117-08:002012-02-19T07:00:58.117-08:00NOTE : X is the point where EG intersects MFNOTE : X is the point where EG intersects MFW Fung, J Hanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32522818597889567242012-02-19T01:41:42.620-08:002012-02-19T01:41:42.620-08:00GX*HX = CX*FX = DX* EX = MX* NX
Hence H,M,G,N is ...GX*HX = CX*FX = DX* EX = MX* NX<br /><br />Hence H,M,G,N is concyclic<br /><br />Hence angle XMG = angle XHN = 36W Fung, J Hanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-979976739193021362012-01-02T21:15:57.362-08:002012-01-02T21:15:57.362-08:00Thanks Anonymous.Thanks Anonymous.Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17153369509326108542012-01-02T12:37:05.608-08:002012-01-02T12:37:05.608-08:00Continuing the reasoning of Ajit:
In right triang...Continuing the reasoning of Ajit: <br />In right triangle AMD: AM^2 = AF * AD ---(1)<br />In right triangle AHC: AH^2 = AE * AC ---(2) <br />Angle(ADE)=ACE) = 90–58 = 32 degree. <br />Then ΔADE~ΔAFC ---> AF/AE = AC/AD <br />So AF * AD = AE * AC ---(3)<br />By (1),(2)and(3) ---> AM = AH.<br />So AH = AG = AM = AN, <br />therefore the quadrilateral HMGN is cyclic.<br />Then angle (NHG)=(NMG)=36 degrees. <br />So x = 36 degrees.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79696675092078223682011-12-31T21:26:53.918-08:002011-12-31T21:26:53.918-08:00Happy New Year to all - I've a suggestion for ...Happy New Year to all - I've a suggestion for Peter, Pravin and others. We've in our diagram E as the midpoint of GH and similarly F is the midpoint of MN. Thus, AH=AG and AM=AN.<br />What I cannot prove is that AG=AM & in fact AG=AH=AM=AN. In other words, a circle with centre at A and radius = AN will pass thru N,G,M & H and thus /_X=36 since/_NHG=36.<br />Now there remains a small matter of proving AG=AM where your expertise is solicited.<br />Cheerio<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com