Every ${K}_{1, 3}$-free connected graph of even order has a perfect matching. 
Every ${K}_{1, 3}$-free connected graph of even order has a perfect matching.

Probably it can be shown using the Tutte theorem about perfect matching.
 A: We want to prove that if $G$ has no perfect matching, then it contains the claw.  Suppose $G$ has no perfect matching. A strengthening of Tutte's theorem, the Gallai-Edmonds theorem, allows us to take $S \subset G$ such that:


*

*$o(G \setminus S) - |S|$ be maximal and

*$S$ is matchable to the components of $G \setminus S$.


We will induct on the order of $S$.
If $|S| = 1$, then $o(G \setminus S) \ge 3$, since $o(G \setminus S) > |S|$ and $o(G \setminus S)$ must be odd since $|G|$ is even. There is an edge from $v \in S$ to all $3$ components, thus providing an example of the claw.
Now suppose $|S| = n+1$ and that the statement holds for $n$.  Select $v_1 \in S$ adjacent to exactly two odd components $C_1$ and $C_2$.  If it is adjacent to $3$, then we have an example of a claw.  $G$ is connected, so we may find $v_2$ with an edge to one of $C_1$ and $C_2$ (let $v_2 c$ be an edge, with $c \in C_1$, without loss of generality).
Now, let $S^* = S - v_1$.  We will show that $S^*$ satisfies our induction hypothesis:


*

*$o(G \setminus S^*) - |S^*|$ is still maximal: $C_1 \cup \{ v_1 \} \cup C_2$ is an odd component of $G \setminus S^*$ and all other components are preserved.

*Moreover, $S^*$ is still matchable to $G \setminus S^*$, since $v_2 c \in E$.

