Question

In triangle ABC EF II AB. ar(CEF) = ar(EFBA). Find CF/EA

Answer 1

Draw a perpendicular from C onto AB to meet AB at G and EF at H.

Area ΔABC = Δ = 1/2 * AB * CG 
  = ΔEFBA + Δ CEF 
  = 2 * Δ CEF = EF * CH
=>  AB * CG =  2 EF * CH   --- (1)

ΔABC and Δ CEF are similar triangles.  So the ratios of their corresponding sides and altitudes are equal.

   EF/AB = CE / CA = CF / CB = CH / CG
  => CH = EF * CG / AB

 So (1) => AB * CG = 2 EF² * CG / AB
=> AB² = 2 EF²
=> AB = √2 EF
The ratio of the sides in the two triangles is √2.

So CE / CA = 1/√2  
 CF / CB = 1/√2

CF / EA  = (CB/√2 ) / [(1 - CA/√2)]
            = CB / [(√2  - 1) AC]