Let .

This problem can done by using the method of symmetry.

(If A and B are connected to battery and AB or it's perpendicular bisector is a line of symmetry, then all points lying on the perpendiculars drawn to AB are at same potential).

 

In the question, the perpendicular to AB is a line of symmetry. Hence the current through AF is equal to the current throguh FB which means no corrent flows from F to C or D. This implies that even if we break the connection at F, the currents in the circuit will not be affected. : Note that I have drawn the circuit only for the topmost layer of triangles.

 

 

Now calculate the resistance between CD.

 

 

 

Now simplifying the circuit, it consists of AB parallel to another branch (of resistance( ) .

Hence,

 

Now for the next layer of triangles, repeat the above procedure to see that



Now we see a pattern. For any particular layer is the resistance is , the next layer the resistance is .

 

So in effect the circuit is this:

 

Hence,

 

   ohms

 

where .