Hi guyz

I think this method is better.

Let us divide line AB of length a into 3 parts by 2 points say I and J.

let AI=x,IJ=y den JB=a-(x+y)

x+y<a

 

x+y>a-(x+y) 

x+y>a/2

 

x+(a-x-y)>y

y<a/2

 

similarly x<a/2

 

Now just sketch these inequalities on a graph

 

Total possible cases are represented by the area of the big triangle formed by x+y<a and the axes

and favorable cases by

the area of the triangle formed by x+y>a/2 x<a/2 and y<a/2

so Probability of formin a triangle =

[a^2/8]/[a^2/2]

gives 1/4