Consider for the case $i >= j$
Push A for each a
Pop A for each b
Push C for each c
Pop C first and then A for each d.
Accept if stack is empty when input finishes.
For the case $i < j$
Push A for each a
Pop A for each b until stack becomes empty
Push B for each b after that
Pop B for each c until stack becomes empty
Push C for each c after that
Pop C first and then B for each d.
Accept if stack is empty when input finishes.
So, clearly $L$ is CFL. But we can combine the two cases and there is no guess needed making $L$ a DCFL.