?
Let the location at which R1 , R4 and R8 are connected together be B ,
and the location at which R3 , R6 and R9 are connected together be C .
R1 = R3 , R4 = R6 and R8 = R9 , so the diagram is bilateral symmetry .
It means that the voltage at B and C are the same .
Therefore , B and C can be shorted without changing the condition .
If we do so , then the diagram becomes a bridge connection as below .
(Ri//Rj means the parallel connection of Ri and Rj .)
.... ------- 10V --------
.... | ........... ........... |
R1//R3 ................ R2
.... | ........... ........... |
.. BC --- R8//R9 --- A
.... | ........... ........... |
R4//R6 ................ R5
.... | ........... ........... |
.... -------- 0V ---------
Let the current from 10V to BC be i1 ,
let the current from 10V to A be i2 , and
let the current from A to BC be i3 .
The voltage difference BC to 10V = (R1//R3)*i1
The voltage difference A to 10V = R2*i2
The voltage difference BC to A = (R8//R9)*i3
So (R1//R3)*i1 = R2*i2 + (R8//R9)*i3 ---(#1)
The voltage difference 0V to BC = (R4//R6)*(i1 + i3)
So (R4//R6)*(i1 + i3) + (R1//R3)*i1 = 10 ---(#2)
The voltage difference 0V to A = R5*(i2 - i3)
So R5*(i2 - i3) + R2*i2 = 10 ---(#3)
Solve the simultaneous equations (#1) to (#3) , and find the voltage difference 0V to A .
That is left for you .