combinatorics – Perfect matchings and equal edges

Let $G=left(V,Eright)$ be a graph and let $M_1,M_2subset E$ be two maximum matchings in G. Show that any path in G whose edges alternate between $M_1$ and $M_2$ and is maximal with respect
to this property must contain equal number of edges from $M_1$ and $M_2$.

What I have done so far,
Because $M_1$ and $M_2$ are maximum matchings in $G$ by berges theorem the graph $G$ contains no M-augmenting paths. Also observe that any path in $G$ can have at most degree 2, if we had 3 edges incident to a given vertex the third edge would either belong to $M_1$ or $M_2$, this is not allowed. Consider a bipartite graph between a vertex set $V_1$ and $V_2$ where n odd $in V_1$ and n even $in V_2$. If we had $left|M_1right|>left|M_2right|$ edges in G would that not create an M-augmenting path and the path would not be maximum so $left|M_1right|=left|M_2right|$. Can someone explain this to me Intuitively as I am new to this.