tag:blogger.com,1999:blog-6933544261975483399.post4558092123098850824..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 900: Intersecting Circles, Common External Tangent, Secant, Angle, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-50919216849988389962016-12-02T01:24:55.296-08:002016-12-02T01:24:55.296-08:00Problem 900
In the convex quadrilateral GCADFE ...Problem 900<br />In the convex quadrilateral GCADFE by Miquel’s theorem the point B is point Miquel.So the F,G,D and B are concyclic.So <CBA=<CFA=<GFD=<GBD.<br />MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27231877611012081072013-07-09T16:44:05.447-07:002013-07-09T16:44:05.447-07:00Since ABFC concyclic, and also BDGF (Problem 899)....Since ABFC concyclic, and also BDGF (Problem 899). <br /><br />Thus <br />∠ABC = ∠AFC = ∠DFG = ∠DBG. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57601670774625303342013-07-09T16:20:07.256-07:002013-07-09T16:20:07.256-07:00Quadri. FCAB is cyclic => ∠ (CBA)= ∠(CFA)
Quadr...Quadri. FCAB is cyclic => ∠ (CBA)= ∠(CFA)<br />Quadr. FGDB is cyclic per result of problem 899=> ∠(DBG)= ∠(GFD)= ∠(CFA)<br />So ∠(ABC)= ∠(DBG)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75220203009722158172013-07-09T15:58:02.265-07:002013-07-09T15:58:02.265-07:00By problem 899, F,B,D,G concylic,
∠GBD=∠GFD=∠CFA=...By problem 899, F,B,D,G concylic, <br />∠GBD=∠GFD=∠CFA=∠CBA<br />nakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.com