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
3rd Solution
DeleteLet < CGB = < CBG = < CFG = < GDA = u..... (1)
So CG is tangent to circle EFG at G and so
CG^2 = CE.CF = 4 X (4 + 6) = 40 ....... (2)
Now <DAG + < GDA = < BGD = < CGD + < CGB
So from (1) < DAG + u = < CGD + u
Hence < DAG = < CGD
So CG is tangential to circle AGC at G and so
CG^2 = CD.CA = 3 X (3 + x) ....(3)
From (2) and (3),
3(3+x) = 40 and
x = 31/3 as before
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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