Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 658.
Thursday, August 18, 2011
Problem 658: Triangle, Altitudes, Diameter, Circle, Angles
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AF^2 = AD.AC
ReplyDeleteAG^2 = AE.AB
But AD.AC = AE.AB since B,E,D,C are concyclic
Follows AF = AG, angles AFG = AGF = x
Now CF^2 = CD.CA = CG.CH
Follows angles CFG = CHF = y say
Hence x + y = AFG + CFG = AFC = 90 deg
Therefore x = 90 - y = 90 - 12 = 78 deg
Why do you state that CD.CA = CG.CH ?
ReplyDeleteRE: Pravin
ReplyDeleteAfter 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
L.p. x= 178
ReplyDelete