Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Consider Tr.ABC , lets assume Angle ACB = x and Angle BAC =y ,We get Angle ABC = 180 - (x+y) Tr. AFB and Tr. BMC are similarAF/FB = BM/CM Angle AFB = Angle BMC = 180 -x - y Now consider Tr.AFG and Tr. NMC AF/FG = MN/CM, angle AFG = angle NMC = x + y , Angle AGF = Angle MCN , by AA similarity Tr.AFG is similar to Tr. APCAngle APC = Angle AFG = x + y Quad. ABCP is cyclic hence P lies on circle O. QED