# 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.