tag:blogger.com,1999:blog-6933544261975483399.post5005107635816390923..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 338. Triangle, Circumcircle, Inscribed Circle, Exterior angle bisector, Concyclic pointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-61712873506865464222016-10-17T22:54:39.731-07:002016-10-17T22:54:39.731-07:00Let the common tangent at E meet AC extended at K....Let the common tangent at E meet AC extended at K.<br /><br />Let < CKE = 2x so that < FEK = < EFK = 90 - x and let < CEK = y = < CBE.<br /><br />So < ACE = < ABE = 2x + y and < ABC = 2x + 2y, hence á = 90 - x – y ….(1)<br /><br />Further < CEF = (90 – x) – y = á = < ABG from (1)<br /><br />Therefore < EBG = < ABG + < ABE = < CEF + < ACE = < AFE proving that<br />GBFE is Concyclic <br /><br />(Note that á = alpha)<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38014018091899756152010-12-04T17:57:49.896-08:002010-12-04T17:57:49.896-08:00http://img34.imageshack.us/img34/1809/problem338.p...http://img34.imageshack.us/img34/1809/problem338.png<br /><br />Connect EA, EB, EDO, DE and EC<br />Draw a line from O perpendicular to AC. This line intersect circle O at point M.<br />1. Tri OME and DFE are isosceles with common side ODE and OM//DF<br />From this we can have m(OEM)=m(DEF) and 3 points E,F and M are collinear .<br />Since M is the midpoint of arc AC so EM become angle bisector of angle AEC<br /><br />2. Note that m(AEC)= m(A) +m(C) and m(AEB)= m(C)<br />So m(BEF)=(m(A)-m(C))/2<br /><br />3. We have m(BGA)=m(A)- alpha and 2.alpha=m(A)+m(C)<br />So m(BGA)= (m(A)-m(C))/2 = m(BEF) and quadrilateral GBFE is concyclic <br /><br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com