Showing that a language satisfies the pumping lemma
Assume that pumping lemma is satisfied for $L$. Let $N$ be the pumping length.

Now take $z = ab^Nc^Nd^N \in L$. $|z| = 3N+1$ and hence this string can be pumped. 

For this string, every choice of $u,v,w,x$ and $y$ that satisfies the conditions of pumping lemma will have a choice of $i \in \mathbb{N}$ such that $uv^iwx^iy \not\in L$. This is a contradiction. Hence $L$ is not context-free.