Prove that if |$V(G)$| is even then $\alpha'(G)=\frac {V(G)}2$

Question

Prove that for a regular graph G, if G is class $1$ (class $1$ = $\Delta (G)$), then |$V(G)$| is even and $\alpha'(G)=\frac {V(G)}2$, ($\alpha'(G)$ =the size of maximum matching).

I had read this reference Show that $\alpha$ $(G)$ $\le$ $\frac {|V(G)|}2$, but this is not my question. Maybe someone can give me solution, hint or other to prove my problem.

My solution :

$\frac {kn}2 = \frac {2l(2m)}2= l.2m$ edeges

Suppose that $c : E -> {1, . . ., k}$ is a coloring of edges of G.

$\frac{l.2m}k=m$,

So there is some color is used on at least $m$ edges. $G$ has only $2m$ vertice, so two of edges must share a vertex and $c$ therefore can not be a proper coloring.

Answer 1

Say $k = \Delta(G)$, so $G$ is a $k$-regular graph. Since $G$ is class 1, there is a proper edge coloring using $k$ colors. Since every vertex is incident to $k$ edges, that means that every vertex is incident to exactly one edge of each color.

So, the set of edges with any given color form a perfect matching. If there are $m$ edges with the given color, then there are $2m$ vertices.