Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, August 13, 2016
Geometry Problem 1244: Circle, Radius, Perpendicular, Chord, Secant, Measurement. Mind Map
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https://goo.gl/photos/FGTGW9iSkgxRKfx16
ReplyDeleteConnect BF, BC and BD
Note that C is the midpoint of arc BG
Triangle BEC similar to FBC ( case AA)
So CB^2=CE.CF
Triangle DCB similar to BCA ( case AA)
So CB^2=CD.CA
And CE.CF=CD.CA or 4x 10= 3(x+3)
So x= 31/3
Let AB cut the circle at G.
ReplyDeleteIn isoceles Tr. BCG,
CG^2 - CE^2 = BE.GE = CE.EF = 24
So CG^2 = 24 + 16 = 40
Now < CGB = CBG = < GDA
Hence < CGD = < DGB - < CGB = < DGB - < GDA = < GAD
Hence CG^2 = CD.CA
So 40 = 3(3+x) from whence x = 31/3
Sumith Peiris
Moratuwa
Sri Lanka
2nd solution
ReplyDelete< CFD = < CGD = < EAD as before
Hence AFED is cyclic and the result is easily calculated
Sumith Peiris
Moratuwa
Sri Lanka
Join points F.O.D then WE CAN eaisly solve for r.
ReplyDeleteNote that FO=OD=OC=r so Triangle FCD is a right triangle
By similar angle WE have traingle FCD is similar to ECA so
4/3+x=3/10 then X=31/3
First you will need to prove the collinearity of F.O.D
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