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