The diode in CD wire is in reverse bias. Hence i3=0
To solve the rest of the circuit we will replace the diodes with 25 ohms resistances.
The simplified version of the circuit becomes,
Now, as we can see the resistances in the branch AB and EF are the same. Hence they will flow equal amount of current.
So, i2=i4. Also, we can see at E that i1=i2+i4=2i2
Using KVL in EFGH we get,
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