Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 658.
AF^2 = AD.ACAG^2 = AE.ABBut AD.AC = AE.AB since B,E,D,C are concyclicFollows AF = AG, angles AFG = AGF = xNow CF^2 = CD.CA = CG.CH Follows angles CFG = CHF = y sayHence x + y = AFG + CFG = AFC = 90 degTherefore x = 90 - y = 90 - 12 = 78 deg
Why do you state that CD.CA = CG.CH ?
RE: PravinAfter you get AF=AG, by symmetry, AH=AG, therefore, H,F,G are on circle (A,AF),so, \angle FAG=2\angle FHG=24deg,so, \angle AGF=\angle AFG=(180-24)/2=78 deg