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Geometry ProblemClick the figure below to see the complete problem 607.
Isosceles triangles ACF, BCE are similar.CF/CE = AC/BCCF.CB = CE.CA, and so A,B,F,E are concyclic.GBA, GFE are secants from G to the circle BAEF.GB.GA = GF.GE,x(x+10)= (8).(12), (x + 5)² = 121,x+5=11, x=6
< BEC = < BCE = < AFC Hence BAEF is cyclicSo 8 x 12 = x(x+10)Solving the quadratic for positive xx = 6Sumith PeirisMoratuwaSri Lanka
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Isosceles triangles ACF, BCE are similar.
ReplyDeleteCF/CE = AC/BC
CF.CB = CE.CA, and so A,B,F,E are concyclic.
GBA, GFE are secants from G to the circle BAEF.
GB.GA = GF.GE,
x(x+10)= (8).(12),
(x + 5)² = 121,
x+5=11, x=6
< BEC = < BCE = < AFC
ReplyDeleteHence BAEF is cyclic
So 8 x 12 = x(x+10)
Solving the quadratic for positive x
x = 6
Sumith Peiris
Moratuwa
Sri Lanka