Triangle, Hard from every angle!!

Construction: Through D, draw a line parallel to AB cutting AE at G and BE at H. Now, through C, draw a line parallel to BA cutting AE at I and ED at J.

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Solution:

In ∆EFB

FD= $\frac{1}{2}$ ED. (Basic Proportionality Theorem)

In ∆CAB

CH=BH= $\frac{1}{2}$ BC (Mid-Point Theorem)

=> HC=EC

In ∆EDH

EJ=DJ= $\frac{1}{2}$ ED. (Mid-Point Theorem)

=> DJ=DF

=> ∆AFD is congruent to ∆CJD

=> AF=JC=a

In ∆AEF

IJ= $\frac{1}{3}$ AF. (Basic Proportionality Theorem)

c= $\frac{1}{3}$ a

In ∆AEB

CI= $\frac{1}{3}$ BA= $\frac{b}{3}$ . (Basic Proportionality Theorem)

$\frac{b}{3}$ =c+a (CI=a+c)

$\frac{b}{3}$ = $\frac{a}{3}$ +a

a= $\frac{b}{4}$

$\frac{b}{a}$ =4

Use Mass Point Geometry.