tag:blogger.com,1999:blog-6933544261975483399.post8038963985616337870..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1141: Inscribed Pentagon, Circle, Triangle, Congruence, Midpoint, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-60413507302852924522015-07-24T23:17:24.843-07:002015-07-24T23:17:24.843-07:00Consider Tr.ABC , lets assume Angle ACB = x and An...Consider Tr.ABC , lets assume Angle ACB = x and Angle BAC =y ,<br />We get Angle ABC = 180 - (x+y) <br /><br />Tr. AFB and Tr. BMC are similar<br />AF/FB = BM/CM <br />Angle AFB = Angle BMC = 180 -x - y<br /> <br />Now consider Tr.AFG and Tr. NMC<br /> AF/FG = MN/CM, angle AFG = angle NMC = x + y , <br />Angle AGF = Angle MCN , by AA similarity Tr.AFG is similar to Tr. APC<br />Angle APC = Angle AFG = x + y<br /> Quad. ABCP is cyclic hence P lies on circle O.<br /> QEDPradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com